Inertial forces in a mechanical system. Inertia force formula

Having established that individual points in Newtonian absolute space are not physical reality, we must now ask: what remains within the framework

this concept at all? The following remains: the resistance of all bodies to acceleration must be interpreted in the Newtonian sense as the action of absolute space. The locomotive that sets the train in motion overcomes the resistance of inertia. A projectile that demolishes a wall derives its destructive power from inertia. The action of inertia occurs whenever accelerations occur, and the latter are nothing more than changes in speed in absolute space (we can use the last expression, since the change in speed has the same magnitude in all inertial systems). Thus, coordinate systems that themselves move with acceleration relative to inertial systems are not equivalent to the latter or to each other. It is possible, of course, to determine the laws of mechanics in such systems, but they will acquire more complex shape. Even the trajectory free body turns out to be no longer uniform and not rectilinear in the accelerated system (see Chapter p. 59). The latter can be expressed in the form of a statement that in an accelerated system, in addition to real forces, there are apparent, or inertial, forces. A body that is not acted upon by actual forces is still subject to the action of these inertial forces, therefore its motion is general case turns out to be uneven and non-linear. For example, a car that starts to move or brakes represents such an accelerated system. Everyone knows the jolt of a train starting or stopping; this is nothing more than the action of the inertial force we are talking about.

Let us consider this phenomenon in detail using the example of a system moving rectilinearly with acceleration. If we measure the acceleration of a body relative to such a moving system, then its acceleration relative to absolute space will obviously be greater by Therefore, the fundamental law of mechanics in this space has the form

If we write it in the form

then we can say that in the accelerated system the law of motion in Newtonian form is satisfied, namely

except that now you need to put K as the force, which is equal to

where K is the actual force, and is the apparent force, or the force of inertia.

So, this force acts on a free body. Its action can be illustrated with the following reasoning: we know that gravity on Earth - the force of gravity - is determined by the formula G = mg, where constant acceleration, due to gravity. The force of inertia acts in this case like gravity; The minus sign means that the inertial force is directed opposite to the acceleration of the reference system that is used as a basis. The magnitude of the visible gravitational acceleration y coincides with the acceleration of the reference frame. Thus, the motion of a free body in the frame is simply motion of the type that we know as falling or the motion of a thrown body.

This relationship between inertial forces in accelerated systems and the force of gravity still seems somewhat artificial here. In fact, it went unnoticed for two hundred years. However, already at this stage we must point out that it forms the basis of Einstein's general theory relativity.

When studying the question of what force of inertia (SI) is, misunderstandings often occur, leading to pseudoscientific discoveries and paradoxes. Let's figure it out this issue, applying scientific approach and justifying everything said with supporting formulas.

The force of inertia surrounds us everywhere. People noticed its manifestations in ancient times, but could not explain it. It was seriously studied by Galileo, and then famous It was because of his extensive interpretation that erroneous hypotheses became possible. This is quite natural, because the scientist made an assumption, and the knowledge accumulated by science in this area did not yet exist.

Newton argued that the natural property of all material objects is the ability to be in a state in a straight line or at rest, provided that they do not turn out to be external influence.

Let's based on modern knowledge Let’s “expand” this assumption. Galileo Galilei also noticed that the force of inertia is directly related to gravity (attraction). And natural attracting objects, the influence of which is obvious, are planets and stars (due to their mass). And since they have the shape of a ball, this is what Galileo pointed out. However, Newton this moment completely ignored.

It is now known that the entire Universe is permeated with gravitational lines of varying intensity. The existence of gravitational radiation is indirectly confirmed, although not mathematically proven. Consequently, the force of inertia always arises with the participation of gravity. Newton also did not take this into account in his assumption of a “natural property”.

It is more correct to proceed from another definition - the indicated force is the value of which is the product of the mass (m) of the moving body and its acceleration (a). The vector is directed counter to the acceleration, that is:

where F, a are the values ​​of the force vectors and the resulting acceleration; m - mass of a moving body (or mathematical

Physics and mechanics offer two names for such an effect: Coriolis and transfer inertial force (PTI). Both terms are equivalent. The difference is that the first option is generally accepted and is used in the mechanics course. In other words, the equality is true:

F kor = F per = m*(-a kor) = m*(-a per),

where F is the Coriolis force; F per - portable inertia force; a kor and a per are the corresponding acceleration vectors.

PSI includes three components: inertia, translational SI and rotational. If there are usually no difficulties with the first, then the other two require clarification. The translational force of inertia is determined by the acceleration of the entire system as a whole relative to any inertial system during a translational type of motion. Accordingly, the third component arises due to the acceleration that appears during the rotation of the body. At the same time, these three forces can exist independently, not being part of the PSI. They are all represented by the same basic formula F = m*a, and the differences are only in the type of acceleration, which, in turn, depends on the type of movement. Thus, they are a special case of inertia. Each of them participates in the calculation of the theoretical absolute acceleration material body (point) in a fixed frame of reference (invisible for observation from a non-inertial frame).

PSI is necessary when studying the issue relative motion, since to create formulas for the motion of a body in a non-inertial system it is necessary to take into account not only other known forces, but also her (F kor or F per).

A non-inertial reference system is a system moving at an accelerated rate relative to an inertial one.

Newton's laws are valid only in inertial frames of reference. Therefore, all the issues considered so far related to inertial systems. However, in practice we often have to deal with non-inertial reference systems. Let us find out how the basic law of dynamics should be written in such systems. Let us first consider the motion of a material point in an inertial reference frame:

Let's introduce no other than her inertial system reference and agree to call the first stationary and the second mobile:

Based on the acceleration addition theorem:

From here we rewrite:

We see that in a non-inertial reference frame the acceleration of a point is determined not only by the force and mass m, but also by the nature of the movement of the moving reference frame itself.

– fictitious forces (they are not caused by the interaction of bodies, but are associated with the accelerated motion of a non-inertial system relative to an inertial one) or inertial forces.

In inertial reference systems, the only reason for the accelerated motion of a material point is the forces acting from material bodies. In non-inertial systems, the cause of accelerated motion is also inertial forces that are not associated with any interaction.

It must be emphasized that inertial forces have a real effect on a point located in a moving coordinate system, since they are included in the equation of motion. Example: the movement of a person in a carriage, when the carriage is moving at a constant speed.

,

.

Now let the car slow down:

.

Thus, the introduction of inertial forces leads to a convenient formulation of the basic laws of mechanics in relative motion and gives them some clarity.

Let's consider two special cases.

Let a material point perform uniform rectilinear motion relative to a moving coordinate system, then, taking into account
we get:

.

Thus, real forces are balanced by the forces of inertia.

Let the material point be at rest with respect to the moving coordinate system:

Then
,

As already noted, Newton's laws are satisfied only in inertial frames of reference. Frames of reference moving relative to an inertial frame with acceleration are called nnon-inertial. In non-inertial systems, Newton's laws, generally speaking, are no longer valid. However, the laws of dynamics can also be applied to them, if, in addition to the forces caused by the influence of bodies on each other, we introduce into consideration forces of a special kind - the so-called inertia forces.

If we take into account the forces of inertia, then Newton’s second law will be valid for any reference system: the product of the mass of a body and the acceleration in the reference frame under consideration is equal to the sum of all forces acting on a given body (including inertial forces). Inertia forces at the same time, they must be such that, together with the forces , caused by the influence of bodies on each other, they imparted acceleration to the body , which it has in non-inertial frames of reference, i.e.

(1)

Because
(is the acceleration of the body in the inertial frame), then

Inertial forces are caused by the accelerated movement of the reference system relative to the measured system, therefore, in the general case, the following cases of manifestation of these forces must be taken into account:

1) inertial forces during accelerated translational motion of the reference system;

2) inertial forces acting on a body at rest in a rotating reference frame;

3) inertial forces acting on a body moving in a rotating reference frame.

Let's consider these cases.

1. Inertial forces during accelerated translational motion of the reference system. Let a ball of mass T. While the cart is at rest or moving uniformly and in a straight line, the thread holding the ball takes a vertical position and the force of gravity
is balanced by the reaction force of the thread .

If the cart is put into forward motion with acceleration , then the thread will begin to deviate from the vertical back to such an angle α until the resultant force
will not provide a ball acceleration equal to . So the resultant force directed towards the acceleration of the cart and for the steady motion of the ball (the ball now moves along with the cart with acceleration ) is equal to
, where
,T. That is, the greater the acceleration of the trolley, the greater the angle of deviation of the thread from the vertical.

With respect to the reference frame associated with the accelerating cart, the ball is at rest, which is possible if the force , which is nothing more than the force of inertia, since no other forces act on the ball. Thus,

(2)

The manifestation of inertial forces during translational motion is observed in everyday phenomena. For example, when a train picks up speed, a passenger sitting in the direction of the train is pressed against the back of the seat under the influence of inertia. On the contrary, when the train is braking, the inertial force is directed towards the opposite side, and the passenger moves away from the seat back. These forces are especially noticeable when the train suddenly brakes. Inertial forces manifest themselves in overloads that occur during the launch and braking of spacecraft.

2. Inertial forces acting on a body at rest in a rotating reference frame. Let the disk rotate uniformly with angular velocity ω (ω =const) around vertical axis, passing through its center. On the disk, at different distances from the axis of rotation, pendulums are installed (balls with a mass of m). When the pendulums rotate together with the disk, the balls deviate from the vertical by a certain angle.

In an inertial reference frame associated, for example, with the room where the disk is installed, the ball rotates uniformly in a circle of radius R(distance from the center of the rotating ball to the axis of rotation). Consequently, it is acted upon by a force whose modulus is equal to F=mω 2 R and the force is directed perpendicular to the axis of rotation of the disk. It is the resultant force of gravity
and thread tension :
. When the motion of the ball is established, then
, where
,T. i.e. the angles of deflection of the pendulum threads will be greater, the greater longer distance R from the center of the ball to the axis of rotation of the disk and the greater the angular velocity of rotation ω .

With respect to the frame of reference associated with the rotating disk, the ball is at rest, which is possible if the force is balanced by an equal and opposite force directed to it , which is nothing more than the force of inertia, since no other forces act on the ball. Force , called centrifugal force of inertia, is directed horizontally from the axis of rotation of the disk and its module is equal to

F ts =mω 2 R (3)

For example, passengers in moving vehicles when turning, pilots when performing aerobatic maneuvers are subject to the action of centrifugal forces of inertia; centrifugal inertial forces are used in all centrifugal mechanisms: pumps, separators, etc., where they reach enormous values. When designing rapidly rotating machine parts (rotors, airplane propellers, etc.), special measures are taken to balance the centrifugal forces of inertia.

From formula (3) it follows that the centrifugal force of inertia acting on bodies in rotating reference frames in the direction of the radius from the axis of rotation depends on angular velocity rotation ω reference and radius systems R, but does not depend on the speed of bodies relative to rotating reference frames. Consequently, the centrifugal force of inertia acts in rotating frames of reference on all bodies located at a finite distance from the axis of rotation, regardless of whether they are at rest in this frame (as we have assumed so far) or moving relative to it with some speed.

3. Inertial forces acting on a body moving in a rotating frame of reference. Let the ball have a mass T moves at constant speed along the radius of a uniformly rotating disk (). If the disk does not rotate, then the ball, directed along the radius, moves along a radial straight line and hits the point A, if the disk is rotated in the direction indicated by the arrow, then the ball rolls along the curve OB, and its speed relative to the disk changes its direction. This is only possible if the ball is acted upon by a force perpendicular to the speed .

D In order to force the ball to roll along a rotating disk along the radius, we use a rod rigidly fixed along the radius of the disk, on which the ball moves without friction evenly and straightly at speed .

When the ball is deflected, the rod acts on it with some force . Relative to the disk (rotating frame of reference), the ball moves uniformly and rectilinearly, which can be explained by the fact that the force is balanced by the inertia force applied to the ball , perpendicular to the speed . This force is called Coriolis inertial force.

It can be shown that the Coriolis force

(4)

Vector perpendicular to the velocity vectors body and angular velocity of rotation reference system in accordance with the right screw rule.

WITH Coriolis force acts only on bodies moving relative to a rotating reference frame, for example, relative to the Earth. Therefore, the action of these forces explains a number of phenomena observed on Earth. So, if a body moves north in the northern hemisphere, then the Coriolis force acting on it, as follows from expression (4), will be directed to the right with respect to the direction of movement, i.e. the body will deviate slightly to the east. If a body moves south, then the Coriolis force also acts to the right when looking in the direction of movement, i.e. the body will deviate to the west. Therefore, in the northern hemisphere there is a stronger erosion of the right banks of rivers; right rails railway tracks in motion wear out faster than the left ones, etc. Similarly, it can be shown that in the southern hemisphere the Coriolis force acting on moving bodies will be directed to the left with respect to the direction of motion.

Thanks to the Coriolis force, bodies falling on the Earth's surface are deflected to the east (at a latitude of 60° this deviation should be 1 cm when falling from a height of 100 m). The behavior of the Foucault pendulum, which at one time was one of the proofs of the rotation of the Earth, is associated with the Coriolis force. If this force did not exist, then the plane of oscillation of a pendulum swinging near the Earth’s surface would remain unchanged (relative to the Earth). The action of Coriolis forces leads to rotation of the oscillation plane around the vertical direction.

,

where the inertia forces are given by formulas (2) – (4).

Let us once again pay attention to the fact that inertial forces are caused not by the interaction of bodies, but accelerated motion of the reference system . Therefore, they do not obey Newton's third law, since if a force of inertia acts on any body, then there is no opposing force applied to that body. The two basic principles of mechanics, according to which acceleration is always caused by force, and force is always caused by the interaction between bodies, are not simultaneously satisfied in reference systems moving with acceleration.

For any of the bodies located in a non-inertial frame of reference, the inertial forces are external; therefore, there are no closed systems here. This means that in non-inertial reference systems the laws of conservation of momentum, energy and angular momentum are not satisfied. Thus, inertial forces act only in non-inertial systems. In inertial frames of reference such forces do not exist.

The question arises about the “reality” or “fictitiousness” of inertial forces. In Newtonian mechanics, according to which force is the result of the interaction of bodies, inertial forces can be viewed as “fictitious”, “disappearing” in inertial reference systems. However, another interpretation is possible. Since the interactions of bodies are carried out through force fields, inertial forces are considered as impacts that bodies are subjected to from some real force fields, and then they can be considered “real”. Regardless of whether inertial forces are considered "fictitious" or "real", many of the phenomena mentioned above can be explained in terms of inertial forces.

The inertial forces acting on bodies in a non-inertial frame of reference are proportional to their masses and, other things being equal, impart identical accelerations to these bodies. Therefore, in the “field of inertial forces” these bodies move in exactly the same way, if only the initial conditions are the same. The same property is possessed by bodies under the influence of gravitational field forces.

Under some conditions, the forces of inertia and the forces of gravity cannot be distinguished. For example, the movement of bodies in a uniformly accelerated elevator occurs in exactly the same way as in a stationary elevator hanging in a uniform field of gravity. No experiment performed inside an elevator can separate a uniform gravitational field from uniform field inertia forces.

The analogy between gravitational forces and inertial forces underlies the principle of equivalence of gravitational forces and inertial forces (Einstein’s equivalence principle): all physical phenomena in the gravitational field occur in exactly the same way as in the corresponding field of inertial forces, if the strengths of both fields at the corresponding points in space coincide, and other initial conditions for the bodies under consideration are the same. This principle is the basis of the general theory of relativity.

Inertia forces and the basic law of mechanics

Bernikov Vasily Ruslanovich,

engineer.

Preface

Internal forces in some cases are the cause of the appearance external forces, attached to the system , , , . Inertial forces are always external in relation to any moving system of material bodies, , , . The forces of inertia act in the same way as the forces of interaction, they are quite real, they can do work, impart acceleration, , , . With a large number of theoretical prerequisites in mechanics about the possibility of using inertial forces as translational ones when creating structures, they did not lead to a positive result. Only a few well-known designs with low efficiency in using inertial forces can be noted: Tolchin's inertsoid, Frolov's vortex liquid propulsion, Thornson's propulsion. The slow development of inertial propulsors is explained by the lack of fundamental theoretical justification observed effect. Based on the usual classical concepts physical mechanics In this work, a theoretical basis has been created for the use of inertial forces as translational ones.

§1. The fundamental law of mechanics and its consequences.

Let us consider the laws of transformation of forces and accelerations into various systems countdown. Let us choose an arbitrarily stationary inertial reference system and agree to consider the motion relative to it to be absolute. In such a reference system, the basic equation of motion is material point is an equation expressing Newton's second law.

m w abs = F, (1.1)

Where F– the force of interaction between bodies.

A body at rest in a moving reference frame is entrained by the latter in its motion relative to a stationary reference frame. This movement is called portable. The motion of a body relative to a reference system is called relative. The absolute movement of a body consists of its relative and portable movements. In non-inertial reference systems (reference systems moving with acceleration), the acceleration transformation law for forward movement has the following form

w abs = w rel +w lane (1.2)

Taking into account (1.1) for forces, we write the equation of relative motion for a material point in a reference frame moving with translational acceleration

mw rel = F - mw lane, (1.3)

Where mw per is a translational force of inertia that arises not due to the interaction of bodies, but due to the accelerated movement of the reference system. The movement of bodies under the influence of inertial forces is similar to movement in external force fields [2, p.359]. The momentum of the center of mass of the system [3, p. 198] can be changed by changing the internal rotational impulse or internal translational impulse. Inertia forces are always external [2, p.359] in relation to any moving system of material bodies.

Let us now assume that the reference system moves completely arbitrarily relative to the stationary reference system. This movement can be divided into two: forward movement with speed v O, equal speed motion of the origin, and rotational motion around the instantaneous axis passing through this origin. Let us denote the angular velocity of this rotation w, and the distance from the origin of the moving reference system to the moving point in it through r. In addition, a moving point has a speed relative to the moving reference frame v rel. Then for absolute acceleration [2, p.362] the relation is known

w abs = w rel - 2[ v rel w] + (d v o /dt) - w 2 r ^ + [ (d w/ dt) r] ,. (1.4)

Where r ^ - radius vector component r, perpendicular to the instantaneous axis of rotation. Let's reschedule relative acceleration to the left side, and the absolute to right side and multiply everything by the mass of the body, we obtain the basic equation of the forces of relative motion [2, p. 364] of a material point in an arbitrarily moving reference frame

mw rel = mw abs + 2m[ v rel w] - m(d v o /dt) + mw 2 r ^ – m[ (d w/ dt) r] . (1.5)

Or accordingly

mw rel = F + F k + F n + F ts + F f, (1.6)

Where: F– force of interaction between bodies; F k – Coriolis inertial force; F n – translational force of inertia; F c – centrifugal force of inertia; F f – phase force of inertia.

Direction of the force of interaction between bodies F coincides with the direction of acceleration of the body. Coriolis inertial force F k is directed according to the vector product of radial and angular velocity, that is, perpendicular to both vectors. Translational force of inertia F n is directed opposite to the acceleration of the body. Centrifugal force of inertia F q is directed along the radius from the center of rotation of the body. Phase inertia force F f is directed opposite to the vector product of angular acceleration and radius from the center of rotation perpendicular to these vectors.

Thus, it is enough to know the magnitude and direction of the forces of inertia and interaction to determine the trajectory of a body’s movement relative to any reference system.

In addition to the forces of inertia and interaction of bodies, there are forces variable mass, which are a consequence of the action of inertial forces. Let's consider Newton's second law in differential form [2, p.77]

d P/dt = ∑ F, (1.7)

Where: P– impulse of the system of bodies; ∑ F– the sum of external forces.

It is known that the momentum of a system of bodies in the general case depends on time and, accordingly, is equal to

P(t) = m(t) v(t), (1.8)

where: m(t) – mass of the system of bodies; v(t) – speed of the system of bodies.

Since velocity is the derivative with respect to time of the coordinates of the system, then

v(t) = d r(t)/dt, (1.9)

Where r– radius vector.

In what follows we will assume the dependence on time of mass, velocity and radius vector. Substituting (1.9) and (1.8) into (1.7) we get

d(m(d r/dt))/dt = ∑ F. (1.10)

Let’s enter the mass m under the differential sign [1, p.295], then

d [ (d(m r)/dt) – r(dm/dt)]/dt = ∑ F.

The derivative of the difference is equal to the difference of the derivatives

d [ (d(m r)/dt) ] dt – d [ r(dm/dt) ] /dt =∑ F.

Let us carry out a detailed differentiation of each term according to the rules of differentiation of products

m(d 2 r/dt 2) + (dm/dt)(d r/dt) + (dm/dt)(d r/dt) +

+ r(d 2 m/dt 2) – r(d 2 m/dt 2)- (dm/dt)(d r/dt) = ∑ F. (1.11)

Let's bring similar members and write equation (1.11) in the following form

m(d 2 r/dt 2) = ∑ F- (dm/dt)(d r/dt). (1.12)

On the right side of equation (1.12) is the sum of all external forces. The last term is called the force of the variable mass, that is

F pm = - (dm/dt)(d r/dt). (1.13)

Thus, another external force is added to the external forces - the force of variable mass. The expression in the first bracket on the right side of equation (1.13) is the rate of change of mass, and the expression in the second bracket is the rate of separation (attachment) of particles. Thus, this force acts when the mass (reactive force) [2, p. 120] of a system of bodies changes with the separation (attachment) of particles with the corresponding speed relative to this system of bodies. Equation (1.12) is the Meshchersky equation [2, p.120], the minus sign indicates that the equation was derived under the assumption of the action internal forces(particle separation). Since equation (1.12) was derived under the assumption that the momentum of a system of bodies changes under the influence of internal forces generating external ones, the exact mathematical method, therefore, when it was derived, two more forces appeared in expression (1.11), which do not participate in changing the momentum of the system of bodies, since they are reduced when similar terms are added. Let us rewrite equation (1.11), taking into account equation (1.13), without canceling similar terms, as follows

m(d 2 r/dt 2) + r(d 2 m/dt 2) +(dm/dt)(d r/dt) = ∑ F + F pm + r(d 2 m/dt 2) +(dm/dt)(d r/dt). (1.14)

Let us denote the penultimate term of expression (1.14) by F m , and the last one through F d, then

m(d 2 r/dt 2) + r(d 2 m/dt 2) + (dm/dt)(d r/dt) = ∑ F + F pm + F m+ F d. (1.15)

Since the strength F m does not participate in the change in momentum, then it can be written as a separate equation

F m = r(d 2 m/dt 2). (1.16)

Let us consider the physical meaning of equation (1.16), for this we rewrite it in the following form

r = F m/(d 2 m/dt 2). (1.17)

The ratio of force to the accelerated growth of mass in a certain volume is a constant value, or the space occupied by a certain amount of a type of substance is characterized by a minimum volume. Force F m is static and performs the function of pressure.

Force F d also does not participate in the change in the momentum of the system of bodies, so let’s write it as a separate equation and consider its physical meaning

F d = (dm/dt)(d r/dt). (1.18)

Force F d is the pressure force exerted by a substance in a liquid or gaseous state to the surrounding space. It is characterized by the number, mass and speed of particles that provide pressure in a certain direction. It should be noted that the pressure force F d coincides with the force of the variable mass F PM and their differentiation is made only to determine the nature of the action in different conditions. Thus, equation (1.15) completely describes the state of matter. That is, considering equation (1.15), we can conclude that a substance is characterized by mass as a measure of inertia, the minimum space that a given amount of substance can occupy without changing its properties, and the pressure exerted by the substance in the liquid and gaseous state on the surrounding space.

§2. Characteristics of the action of inertial forces and variable mass.

Translational accelerated motion of a body occurs under the influence of force according to Newton's second law. That is, a change in the magnitude of the speed of a body occurs in the presence of acceleration and the force that caused this acceleration.

The use of centrifugal inertial force for translational motion is possible only with an increase in the linear speed of the sources of these forces, since with the accelerated movement of the system, the inertial forces of the sources in the direction of increasing the speed of the system decrease until they disappear completely. In addition, the field of inertial forces must be non-uniform and have maximum value in the part of the system in the direction of translational movement.

Consider the movement of a body (Fig. 2.1) with mass m in a circle of radius R.

Rice. 2.1.

Centrifugal force Fμ with which the body presses on the circle is determined by the formula

F q = m ω 2 R. (2.1)

Using the known relation ω = v /R, where v is the linear velocity of the body perpendicular to the radius R, we write formula (2.1) in the following form

F c = m v 2 / R. (2.2)

Centrifugal force acts in the direction of the radius R. Now let’s instantly break the circle along which the body is moving. Experience shows that the body will fly tangentially in the direction of linear velocity v, and not in the direction of the centrifugal force. That is, in the absence of support, the centrifugal force instantly disappears.

Let a body of mass m move along an element of a semicircle (Fig. 2.2) with radius R, and the semicircle moves with acceleration w П perpendicular to the diameter.

Rice. 2.2.

With uniform motion of the body (the linear speed does not change in magnitude) and an accelerated semicircle, the support in the form of a semicircle instantly disappears and the centrifugal force will be equal to zero. If a body moves with positive linear acceleration, then it will catch up with the semicircle and the centrifugal force will act. Let's find the linear acceleration w of the body at which the centrifugal force acts, that is, presses on the semicircle. To do this, the time spent by the body on a tangential path until it intersects with a dashed line parallel to the diameter and drawn through point B (Fig. 2.2) must be less than or equal to the time spent by the semicircle in the direction perpendicular to the diameter. Let the initial velocities of the body and the semicircle be equal to zero and the elapsed time be the same, then the path S AC traversed by the body

S AC = w t 2 /2, (2.3)

and the path traversed by semicircle S AB will be

S AB = w P t 2 /2. (2.4)

Dividing equation (2.3) by (2.4) we get

S AC / S AB = w / w P.

Then the acceleration of the body w, taking into account the obvious relation S AC / S AB = 1/ cosΨ

w = w П /cosΨ, (2.5)

where 0 £ Ψ £ π/2.

Thus, the projection of the body’s acceleration in a circle element in a given direction (Fig. 2.2) must always be greater than or equal to the acceleration of the system in the same direction in order to maintain centrifugal force in action. That is, centrifugal force acts as a translational force driving force only in the presence of positive acceleration, changing the magnitude of the linear velocity of the body in the system

The relationship for the second quarter of the semicircle is obtained similarly (Fig. 2.3).

Rice. 2.3.

Only the path traversed by the body along a tangent will begin from a point on a semicircle moving with acceleration until it intersects with a dashed line parallel to the diameter and passing through point A initial position semicircles. The angle in this case is determined by the interval π/2 ³ Ψ ³ 0.

For a system in which a body moves uniformly or with deceleration in a circle, the centrifugal force will not cause a translational accelerated motion of the system, since the linear acceleration of the body will be zero or the body will lag behind the accelerated motion of the system.

If a body rotates with angular velocity ω and at the same time approaches the center of the circle at a speed v, then the Coriolis force arises

F k = 2m [ v ω]. (2.6)

A typical trajectory element is shown in Fig. 2.4.

Rice. 2.4.

All formulas (2.3), (2.4), (2.5) and conclusions for maintaining the centrifugal force of the circulating medium in action will also be true for the Coriolis force, since with accelerated motion of the system, a body moving with positive linear acceleration will keep pace with the acceleration of the system and , accordingly, move along curvilinear trajectory, and not along a tangent line, when there is no Coriolis force. The curve must be divided into two halves. In the first half of the curve (Fig. 4), the angle changes from the initial point to the bottom in the interval -π/2 £ Ψ £ π/2, and in the second half from the bottom point to the center of the circle π/2 ³ Ψ ³ 0. Similarly, with rotation of the body and its simultaneous removal (Fig. 2.5) from the center, the Coriolis force acts as translational with a positive acceleration of the linear velocity of the body.

Rice. 2.5.

The interval of angles in the first half from the center of the circle to the bottom point is 0 £ Ψ £ π/2, and in the second half from the bottom point to the final point π/2 ³ Ψ ³ -π/2.

Let us consider the translational force of inertia F n (Fig. 2.6), which is determined by the formula

F n = -m w,(2.7)

Where w– acceleration of the body.

Rice. 2.6.

With positive acceleration of a body, it acts against the movement, and with negative acceleration (deceleration), it acts in the direction of movement of the body. When an element of acceleration or deceleration (Fig. 2.6) acts on the system with which the elements are connected, the acceleration of the element’s body in absolute value must obviously be greater than the module of acceleration of the system caused by the translational force of inertia of the body. That is, the translational force of inertia acts as a driving force in the presence of positive or negative acceleration.

Phase inertia force F f (inertial force caused by uneven rotation) is determined by the formula

Fφ = -m [(d ω /dt) R]. (2.8)

Let the radius R perpendicular to the angular velocity vector ω , then in scalar form formula (2.8) takes the form

F f = -m (dω/dt)R. (2.9)

With positive angular acceleration of the body (Fig. 1.7), it acts against the movement, and with negative angular acceleration (deceleration) it acts in the direction of movement of the body.

Rice. 2.7.

Using the known relation ω = v /R, where v is the linear velocity of the body perpendicular to the radius R, we write formula (2.9) in the following form

F f = -m (dv/dt). (2.10)

Since dv/dt =w, where w is the linear acceleration of the body, then equation (2.10) takes the form

F f = -m w (2.11)

Thus, formula (2.11) is similar to formula (2.7) for the translational inertial force, only the acceleration w must be decomposed into parallel α II and perpendicular α ┴ components (Fig. 2.8) with respect to the diameter of the semicircle element.

Rice. 2.8.

Obviously, the perpendicular component of acceleration w ┴ creates a torque, since in the upper part of the semicircle it is directed to the left, and in the lower part to the right. The parallel component of acceleration w II creates a translational force of inertia F fII, since it is directed in the upper and lower parts of the semicircle in one direction, coinciding with the direction w II.

F fII = -m w II. (2.12)

Using the relation w II = w cosΨ, we obtain

F ФII = -m w cosΨ, (2.13)

where the angle Ψ is in the interval -π/2 £ Ψ £ π/2.

Thus, formula (2.13) is obtained for calculating the element of the phase inertia force for translational motion. That is, the phase inertia force acts as a driving force in the presence of positive or negative linear acceleration.

So, four elements of translational inertial force have been identified: centrifugal, Coriolis, translational, phase. Connecting individual elements in a certain way, it is possible to combine systems of translational driving force of inertia.

Consider the force of a variable mass, defined by the formula

F pm = - (dm/dt)(d r/dt). (2.14)

Since the speed of detachment (attachment) of particles relative to the system of bodies is equal to

u=d r/dt, (2.15)

then we write equation (2.14) as follows

F pm = - u(dm/dt). (2.16)

In equation (2.16), the variable mass force is the value of the force produced by the separating particle as its speed changes from zero to u or the value produced by the joining particle during a change in its speed from u to zero. Thus, the force of variable mass acts at the moment of acceleration or deceleration of particles, that is, it is a translational force of inertia, but calculated according to other parameters. Taking into account what has been written above, there is a need to clarify the derivation of the Tsiolkovsky formula. We rewrite equation (1.12) in scalar form and set ∑ F= 0, then

m(d 2 r/dt 2) = - (dm/dt)(dr/dt). (2.17)

Since the system acceleration

d 2 r/dt 2 = dv/dt,

where v is the speed of the system, then equation (2.17) taking into account equation (2.15) will be

m(dv/dt) = - (dm/dt)u. (2.18)

Multiplying equation (2.17) by dt we get

mdv = -udm, (2.19)

that is, knowing the maximum speed u = u O of particle separation, which we consider constant, we can determine from the ratio of the initial m O and final masses m final speed systems v

v = -u O ∫ dm /m = u O ln(m O /m). (2.20)

m O /m = e v/uo . (2.21)

Equation (2.21) is the Tsiolkovsky equation.

§3. Contour of the circulating medium of centrifugal force of inertia.

Let us consider the circulation of a medium along a torus (Fig. 3.1) with an average radius R, moving with an angular velocity ω relative to the center O . The modulus of the centrifugal force acting on a point flow element with mass ∆m will be equal to

∆ F= ∆m ω 2 R.

In any section of the ring for identical elements the centrifugal force will be the same in magnitude and directed radially from the center, stretching the ring. Centrifugal force does not depend on the direction of rotation.

Rice. 3.1.

Now let's calculate the total centrifugal force acting perpendicular to the diameter of the upper semicircle (Fig. 3.2). Obviously, in the direction from the middle of the diameter perpendicular projection the force will be maximum, gradually decreasing towards the edges of the semicircle, due to the symmetry of the curve relative to the midline. In addition, the resultant of the projections of centrifugal forces acting parallel to the diameter will be equal to zero, since they are equal and opposite in direction.

Rice. 3.2.

Let us write down the elementary function of the centrifugal force acting on a point segment with mass ∆ m and length ∆ ℓ:

∆ F= ∆ m ω 2 R. (3.1)

The mass of a point element is equal to the flux density multiplied by its volume

∆ m=ρ ∆ V. (3.2)

Length of half torus along the midline

where π is the number pi.

Volume of half a torus

V = π 2 Rr 2 = πR π r 2 = ℓ π r 2 ,

where r is the radius of the torus tube.

For an elementary volume we write

∆ V= ∆ ℓ π r 2 .

It is known that for a circle

∆ ℓ= R ∆ Ψ,

∆ V = π r 2 R ∆ Ψ. (3.3)

Substituting expression (3.3) into (3.2) we get:

∆ m=ρ π r 2 R ∆ Ψ. (3.4)

Now let’s substitute (3.4) into (3.1), then

∆ F= ρ π r 2 ω 2 R 2 ∆ Ψ.

Centrifugal force acting in perpendicular direction(Fig.2)

∆ F┴ = ∆ Fcos((π/2)- Ψ).

It is known that cos((π/2)- Ψ)=sin Ψ, then

∆ F┴ = ∆ F sin Ψ.

Let's substitute the value for ∆ F we get

∆ F┴ = ρ π r 2 ω 2 R 2 sin Ψ ∆ Ψ.

Let's find the total centrifugal force acting in the perpendicular direction in the interval from 0 to Ψ

F ┴ = ∫ ρ π r 2 ω 2 R 2 sin ΨdΨ.

Let's integrate this expression, then we get

F ┴ = - ρ π r 2 ω 2 R 2 cosΨ│. (3.5)

Let us assume that the acceleration w of the circulating medium is ten times more acceleration system w c, that is

In this case, according to formula (2.5) we obtain

Let's calculate the angle of action of inertial forces in radians

Ψ ≈ 0.467 π,

which corresponds to an angle of 84 degrees.

Thus, the angular range of action of inertia forces is

0 £ Ψ £ 84° in the left half of the contour and symmetrically 96° £ Ψ £ 180° in the right half of the contour. That is, the absence interval active forces inertia in the entire circuit is about 6.7% (in reality, the acceleration of the circulating medium is much greater than the acceleration of the system, so the interval of absence of acting inertia forces will be less than 1% and can be ignored). To determine the total centrifugal force in these angle intervals, it is enough to substitute the first interval into formula (3.5) and, due to symmetry, multiply by 2 we get

F ┴ = - 2ρ π r 2 ω 2 R 2 cosΨ│. (3.6)

After simple calculations we get

F ┴ = 1.8 ρ π r 2 ω 2 R 2 .

It is known that the angular velocity

F ┴ = 1.8 ρ π r 2 v 2 .

Since the circulating medium must move with acceleration for the inertial force to act, we therefore express the linear speed in terms of acceleration, assuming the initial speed equal to zero

F ┴ = 1.8 ρ π r 2 (w t) 2 . (3.8)

The average value during the positive acceleration, which we consider constant, will be

F ┴CP = ((1.8ρ π r 2 w 2)/t) ∫t 2 dt.

After calculations we get

F ┴SR = 0.6ρ π r 2 w 2 t 2 . (3.9).

Thus, a contour of the circulating medium was identified, from which a closed chain can be formed and their centrifugal forces can be summed up.

Let's make a closed circuit of four contours of different sections (Fig. 3.3): two upper contours of radius R, section S, and two lower contours of radius R1, section S1, neglecting edge effects when the circulating medium passes from one section to another. Let S< S 1 и радиус

R 1< R. Плотность циркулирующей среды одинакова. Тогда согласно уравнению неразрывности отношение скоростей потока в разных сечениях обратно пропорционально их сечениям, то есть

v/v 1 = S 1 /S = r 1 2 /r 2 , (3.10)

where r 1 and r are the radii of the flow of the circulating medium of the corresponding section.

In addition, let us write down the obvious relation for velocities and accelerations

v/v 1 = w/w 1. (3.11)

Let's find the acceleration of the lower contour medium using equation (3.10) and (3.11) for calculations

w 1 = w r 2 / r 1 2 . (3.12)

Now, according to equation (3.9), we determine the centrifugal force for the lower contour, taking into account equation (3.12) and after calculations we obtain

F ┴CP1 = 0.6 ρ π r 1 2 w 1 2 = 0.6ρ π r 2 w 2 t 2 (r 2 / r 1 2) = F ┴CP (r 2 / r 1 2) (3.13)

When comparing the expression for the centrifugal force of the upper contour (3.9) and the lower contour (3.13), it follows that they differ by the amount (r 2 / r 1 2).

That is, when r< r 1 центробежная сила верхнего контура больше, чем нижнего.

Rice. 3.3.

The resultant of centrifugal forces acting on two contours in the upper half-plane (the boundary of the upper and lower half-plane is shown by a thin line) is oppositely directed to the resultant of centrifugal forces acting on two contours in the lower half-plane. Obviously, the total F C centrifugal force will act in the direction as shown in Figure 3.3; let’s take this direction as positive. Let's calculate the total centrifugal force F

F C = 2 F ┴SR - 2F ┴SR1 = 1.2ρ π r 2 w 2 t 2 (1- (r 2 / r 1 2)) (3.14)

As we can see, the total centrifugal force depends on the flow density, cross sections of opposite contours and flow acceleration. The total centrifugal force does not depend on the radius of the contours. For a system in which the circulating medium moves uniformly or with deceleration along the circumference, centrifugal force will not cause progressive accelerated motion of the system.

Thus, the basic contour of the circulating medium was identified, and the possibility of using the contours of the circulating medium of different sections to sum up the centrifugal force in a certain direction and change the total impulse of a closed system of bodies under the influence of external inertial forces caused by internal forces was shown.

Let r = 0.025m; r 1 = 0.05m; ρ = 1000 kg/m 3 ; w = 5m/s 2, t = 1s, then during the positive acceleration the average value total centrifugal force F C.≈ 44N.

§4. Contour of the circulating medium of Coriolis inertial force.

It is known that the Coriolis inertial force arises when a body of mass m rotates in a circle and simultaneously moves radially, and it is perpendicular to the angular velocity ω and radial movement speed v. Direction of Coriolis force F coincides with the direction vector product in the formula F= 2m[ vw].

Rice. 4.1.

Figure 4.1 shows the direction of the Coriolis force when a body rotates in a circle counterclockwise and moves radially towards the center of the circle during the first half-cycle. and Fig. 4.2 shows the direction of the Coriolis force when the body rotates in a circle, also counterclockwise, and radially moves it from the center of the circle during the second half-cycle.

Rice. 4.2.

Let's combine the left part of the body movement in Fig. 4.1 and the right part in Fig. 4.2. then we get in Fig. 4.3 variant of the trajectory of a body’s movement over a period.

Rice. 4.3.

Let's consider the movement of a circulating medium (liquid) through pipes curved according to the trajectory. The Coriolis forces of the left and right curves act in a sector of 180 degrees in the radial direction when moving from point B to point O to the left and right, respectively, relative to the X axis. Components of the Coriolis force of the left and right curves F| | parallel to the straight line AC compensate each other, since they are identical, oppositely directed and symmetrical relative to the X axis. The symmetrical components of the Coriolis force of the left and right curve F^ perpendicular to the straight line AC add up, since they are directed in the same direction.

Let's calculate the magnitude of the Coriolis force acting along the X axis on the left half of the trajectory. Since composing the trajectory equation represents difficult task, then we look for a solution to find the Coriolis force using an approximate method. Let v be the fluid velocity constant along the entire trajectory. Radial speed v r and linear speed of rotation v l, according to the speed parallelogram theorem, we express (Fig. 3) through speed v and angle α

v r = v cosα, v l = v sinα.

The trajectory of movement (Fig. 4.3) is constructed taking into account the fact that at point B the radial velocity v r is equal to zero, and the linear velocity v l is equal to v. At the center of a circle O, with radius Ro, the radial velocity v p is equal to v, and the linear velocity v l is equal to zero, and the tangent trajectory at the center of the circle is perpendicular to the tangent trajectory at the beginning (point B). The radius decreases monotonically from Ro to zero. Angle α changes from 90° at point B to 0° at the center of the circle. Then, from graphical constructions, we select the length of the trajectory 1/4 of the length of the circle with radius R 0. Now you can calculate the mass of the liquid using the formula for the volume of a torus. That is, the mass of the circulating medium will be equal to 1/4 of the mass of the torus with an average radius R 0 and an internal radius of the pipe r

m = ρπ 2 r 2 R 0 /2, (4.1)

where ρ is the density of the liquid.

The modulus of the projection of the Coriolis force at each point of the trajectory onto the X axis is found by the formula

F^ = 2m v р ср ω ср cos b , (4.2)

where v r av – average value of radial velocity; ω av – average value of angular velocity; b is the angle between the Coriolis force F and the X axis (-90° £ b £ 90° ).

For technical calculations, the interval without the action of inertial forces can be ignored, since the acceleration of the circulating medium is much greater than the acceleration of the system. That is, we select the interval of angles between the Coriolis force F and the X axis (-90° £ b £ 90° ). Angle α changes from 90° at point B to 0° at the center of the circle, then the average value of the radial speed

v р ср = 1 / (0 - π/2) ∫ v cos α dα = 2 v / π. (4.3)

The average angular velocity will be equal to

ω av = (1/ ((v π /2Rо) - v Ro))) ∫ ω dω = (v /2Rо) ((π /2.) +1). (4.4)

The lower limit of the angular velocity of the integral in formula (4.4) is determined in starting point B. It is obviously equal to v / Ro. The upper value of the integral is defined as the limit of the ratio

ℓim (v l /R) = ℓim (v sinα /R), (4.5)

v l ® 0 α ® 0

R ® 0 R ® 0

where R is the current radius.

Let's use the well-known method [7, p. 410] of finding limits for functions of several variables: the function vsinα /R at the point (R= 0, α = 0) on any straight line R = kα passing through the origin has a limit. In this case, there is no limit, but there is a limit for a certain line. Let's find the coefficient k in the equation of the straight line passing through the origin.

At α = 0 ® R= 0, at α = π /2 ® R= Ro (Fig. 3), hence to = 2Ro/π, then formula (5) is transformed to a form that includes the first remarkable limit

ℓim (v π sinα /2Ro α) = (v π/2Ro) ℓim sinα/α = v π/2Ro. (4.6)

α ® 0 α ® 0

Now we substitute the obtained value from formulas (4.1), (4.3) and (4.4) into (4.2) and we get

F^ = ρ π r 2 v 2 ((π /2.) +1) cos b .

Let's find the sum of projections of the Coriolis force in the interval (-90° £ b £ 90° ) for the left curve.

90°

F^ = ρ π r 2 v 2 ((π /2.) +1) ∫ cos b db = 2 ρ π r 2 v 2 ((π /2.) +1).

90°

The final sum of the Coriolis force projections for the left and right curves

∑F^ = 4ρ r 2 v 2 ((π /2.) +1). (4.7)

According to relation (3.7), we rewrite equation (4.7) in the form

∑F^ = 4ρ r 2 (w t) 2 ((π /2.) +1). (4.8)

Let us calculate the average value of the Coriolis force over time, considering the acceleration constant

Fк = ∑F^ ср = 4ρ r 2 w 2 ((π /2.) +1) / t) ∫t 2 dt.

After calculations we get

Fк ≈ 1.3ρ r 2 w 2 ((π /2.) +1)t 2 . (4.9)

Let r = 0.02m; w = 5m/s 2 ; ρ = 1000kg/m3; t = 1c, then the total average Coriolis inertial force during the positive acceleration of the circulating medium will be Fк ≈ 33N.

In the center of the circle in the trajectory there is an inflection (Fig. 4.3), which can be interpreted, to simplify calculations, as a semicircle with a small radius. For clarity, let’s divide the trajectory into two halves and insert a semicircle into the lower part, and into top part straight line, as shown in Fig. 4.4, and direct the circulating medium through a pipe of radius r, curved according to the shape of the trajectory.

Rice. 4.4.

In formula (3.5) we set the angle Ψ = 180°, then the total centrifugal force Fc acting in the perpendicular direction for the circuit of the circulating medium

Fts = 2 ρπ r 2 v 2 . (4.10)

Thus, the centrifugal force does not depend on the radius R, but depends only on the angle of integration (see formula (3.5)) at a constant flux density ρ, radius r and velocity of the circulating medium v ​​at each point of the trajectory. Since the radius R can be any, we can conclude that for any convex curve with edges perpendicular to the straight line AOB (Fig. 3.2), the centrifugal force will be determined by expression (4.10). It should be noted as a consequence that each edge of a convex curve can be perpendicular to its line, which are parallel and do not lie on the same line.

The sum of the projections of centrifugal forces (Fig. 4) acting against the direction of the X axis, arising in a semicircle and two halves of a convex curve (the straight line does not contribute to the centrifugal force) above the broken line and projections acting along the X axis, arising in two convex curves under the broken line is compensated, since they are identical and directed in opposite directions. Thus. centrifugal force does not contribute to forward motion.

§5. Solid state rotational systems. Centrifugal forces of inertia.

1. The vector of the rods’ own angular velocity is perpendicular to the angular velocity vector of the center of mass of the rod and the radius of the common axis of rotation of the rods.

The energy of translational motion can be converted into energy rotational movement and vice versa . Consider a pair of opposite rods of length ℓ with point weights of equal mass at the ends, uniformly rotating around their own center of mass and around general center About radius R with angular velocity ω (Fig. 5.1): half a turn of the rod in one revolution around a common axis. Let R³ ℓ/2. For a complete description of the process, it is enough to consider rotation in the angle range 0£ α £ π/2. Let us arrange the forces acting parallel to the X axis passing through the common center O and the position of the rods at an angleα = 45 degrees, in the plane of the X axis and the common axis of rotation, as shown in Figure 5.1.

Rice. 5.1.

Angle α is related to frequency ω and time t by the relation

α = ωt/2, (5.1.1)

since half a revolution of the rod occurs in one revolution around a common axis. It is obvious that centrifugal forces inertia there will be more distant loads from the center than nearby ones. Projections of centrifugal forces inertia on the X axis will be

Ft1 = mω 2 (R - (ℓ/2) cos α) sin 2α (5.1.2)

Ft2 = mω 2 (R + (ℓ/2) cos α) sin 2α (5.1.3)

Ft3 = - mω 2 (R + (ℓ/2) sin α) sin 2α (5.1.4)

Ft4 = - mω 2 (R - (ℓ/2) sin α) sin 2α (5.1.5)

Let us write down the difference centrifugal force inertia , acting on remote loads. Difference centrifugal force inertia for the second load

Ft2-1 = mω 2 ℓ cosα sin2α. (5.1.6)

Difference centrifugal force inertia for the third load

Ft3-4 = - mω 2 ℓ sinα sin2α. (5.1.7)

Average value of difference centrifugal forces inertia for half a turn it will be

Fav ц2-1 = (1/(π/2))∫mω 2 ℓ cosα sin2αdα = 4mω 2 ℓ/3 π » 0.4mω 2 ℓ, (5.1.8)

Fav c3-4 = (1/(π/2))∫mω 2 ℓ sinα sin2αdα = -4mω 2 ℓ/3 π » -0.4mω 2 ℓ. (5.1.9)

We obtained two opposite and equal in magnitude centrifugal forces inertia, which are external. Therefore, they can be represented as two identical bodies at infinity (not included in the system), simultaneously interacting with the system: the second load pulls the system towards the first body, and the third load pushes the system away from the second body.

The average value of the force of forced influence on the system per half-turn along the X axis is equal to the sum of the pulling forces Fav c2-1 and repulsion Fav c3-4 from external bodies

Fп = | Fav c2-1 | + | Fav c3-4 | = 0.8 mω 2 ℓ. (5.1.10)

To eliminate the torque of a system of two rods in a vertical plane (Fig. 5.2), it is necessary to use another pair of opposite rods, rotating synchronously in the same plane in the opposite direction.

Rice. 5.2.

To eliminate the torque of the system along a common axis with center O, we use the same pair of four rods, but rotating in the opposite direction relative to the common axis (Fig. 5.3).

Rice. 5.3.

Finally, for a system of four pairs of rotating rods (Fig. 5.3), the traction force will be

Ft = 4Fp = 3.2mω 2 ℓ. (5.1.11)

Let m = 0.1kg; ω =2 πf, where f = 10 rev/s; ℓ = 0.5m, then Ft ≈ 632N.

2. The vector of the rods’ own angular velocity is perpendicular to the angular velocity vector of the center of mass of the rod and parallel to the radius of the common axis of rotation of the rods.

Let us consider a pair of opposite rods perpendicular to each other of length ℓ with point loads of equal mass at the ends, uniformly rotating around their own center of mass and around a common center O with radius R with angular velocity ω (Fig. 5.4): half a revolution of the rod per revolution around a common axis.

Rice. 5.4.

For calculation we choose only m1 and m2, since the solution is similar for m3 and m4. Let us determine the angular velocities of the loads relative to the common center O. The modules of the projections of the linear speed of the loads relative to their own center of mass parallel to the plane of rotation relative to the common center O will be (Fig. 5.5)

v1 = v2 = (ωℓ/4) sin (Ψ/2), (5.2.1)

where Ψ = ωt.

Let us select by absolute value the projections of the tangent of these velocities perpendicular to the radii r1 and r2 respectively relative to the center O we get

v1R = v2R = (ωℓ/4) sin ( Ψ /2)cosb, (5.2.2)

cosb= R /r1 = R /r2 =R/Ö (R 2 +(ℓ 2 /4) cos 2 (Ψ /2)), (5.2.3)

R – distance from the center O to the center of mass of the loads, r1, r2 – distance from the loads to the center O, and r1 = r2.

Rice. 5.5.

The modules of the linear velocity of the loads relative to the common center O without taking into account their linear velocity relative to their own center of mass will be

vR1 = ω r1, (5.2.4)

vR2 = ω r2. (5.2.5)

Let us find the total angular velocity of each load relative to the common axis of rotation, taking into account that the linear velocities are in opposite directions for the first load and the same for the second, then

ω 1 = (vR1 - v1R)/r1 = ω [ 1– (ℓR sin (Ψ/2))/4(R 2 +(ℓ 2 /4)cos 2 (Ψ/2)) ] , (5.2.6)

ω 2 = (vR2 + v2R)/r2 = ω [ 1+ (ℓR] . (5.2.7)

Accordingly, the centrifugal forces will be

F 1 = mω 1 2 r1

F 2 = mω 2 2 r2

Or in detail

F 1 = mω 2 [ (1– (ℓR sin(Ψ/2))/4(R 2 +(ℓ 2 /4)cos 2 (Ψ/2)) ] 2 Ö (R 2 +(ℓ 2 /4)cos 2 (Ψ/2)), (5.2.8)

F 2 = mω 2 [ (1+ (ℓR sin(Ψ/2))/4(R 2 +(ℓ 2 /4)cos 2 (Ψ/2)) ] 2 Ö (R 2 +(ℓ 2 /4)cos 2 (Ψ/2)). (5.2.9)

Let's consider the option when ℓ= 4R. In this case, whenΨ=180° angular frequency of the first load ω 1 = 0 and it does not change direction, the second load has ω 2 = 2ω (Fig. 5.6).

Rice. 5.6.

Let's move on to determining centrifugal forces in the direction of the X axis at ℓ= 4R

F 1 = mω 2 R [ (1+ 4cos 2 (Ψ/2)– sin(Ψ/2))/(1+4cos 2 (Ψ/2)) ] 2 Ö (1 + 4cos 2 (Ψ/2)), (5.2.10)

F 2 = mω 2 R [ (1+ 4cos 2 (Ψ/2)+ sin(Ψ/2))/(1+4cos 2 (Ψ/2)) ] 2 Ö (1 + 4cos 2 (Ψ/2)). (5.2.11)

It should be noted that with increasing angleΨ from 0 to 180 ° at pointΨ = b = 60 ° projection of centrifugal force F 2 changes sign from negative to positive.

First, we add the average values ​​of the projection onto the X-axis of the centrifugal force of the first load and the average value of the projection of the second in the angle interval

0 £ Ψ £60° , taking into account the signs, since they are oppositely directed

F CP 1-2 = (1/(π /3))∫ (F 1 sin( b+Ψ) - F 2 sin( b-Ψ))dΨ ≈ 0.6mω 2 R, (5.2.12)

Where b = arccos(1/ Ö (1 +4 cos 2 (Ψ /2))) is determined from formula (5.2.3).

Centrifugal force F CP 1-2 in formula (5.2.12) is positive, that is, directed along the X axis. Now we add the equally directed average value of the projection onto the X-axis of the centrifugal force of the first load and the average value of the projection of the second in the angle interval of 60° £ Ψ £180°

F CP 1+2 = (1/(π-(π/3)))∫(F 1 sin(Ψ + b)+ F 2 sin(Ψ- b))dΨ ≈ 1.8mω 2 R, (5.2.13)

Average value in interval 0° £ Ψ £180° obviously there will be

F CP = (F CP 1-2 + 2F CP 1+2)/3 ≈ 1.4 mω 2 R. (5.2.14)

For m3 and m4, the average value of the projection onto the X-axis of the centrifugal force will be the same, but acting in the opposite direction.

F T = 4 F CP = 5.6mω 2 R. (5.2.15)

Let m = 0.1kg; ω =2 πf, where f = 10 rev/s; ℓ= 4R, where R = 0.1m, then F T ≈ 220N.

3. The vector of the rods’ own angular velocity is parallel and identically directed with the angular velocity vector of the center of mass of the rod rotating about a common axis.

Let us consider a pair of opposite rods lying on the water plane of length ℓ with point loads of equal mass at the ends, uniformly rotating around their own center of mass and around a common center O with radius R with angular velocity ω (Fig. 5.7): half a revolution of the rod per revolution around a common axis.

Rice. 5.7.

Similar to the previous case, we select only m1 and m2 for calculation, since the solution is similar for m3 and m4. Approximate estimate we will produce the acting inertial forces at ℓ = 2R using the average values ​​of the angular velocity relative to the center O, as well as the average values ​​of the distance from the loads to the center O. Obviously, the angular velocity of the first load at the beginning will be 1.5ω of the second load 0.5ω, and after half a turn both have ω. The distance from the first weight to the center O at the beginning is 2R from the second weight 0, and after half a turn from each RÖ 2.

Rice. 5.8.

Moreover, in the interval 0° £ Ψ £36° (Fig. 5.8) centrifugal forces add up in the direction of the X axis, in the interval 36° £ Ψ £72° (Fig. 5.8, Fig. 5.9) the force of the second is subtracted from the force of the first body and their difference acts along the X axis, in the interval 72° £ Ψ £90° (Fig. 5.9) the forces add up and act opposite to the X axis.

Rice. 5.9.

Let us determine the average values ​​of the angular velocity and radii of the loads per half-turn.

Average angular velocity of the first load

ω CP 1 = (ω + 0.5ω + ω)/2 = 1.25ω. (5.3.1)

Average angular velocity of the second load

ω CP 2 = (ω - 0.5ω + ω)/2 = 0.75ω. (5.3.2)

Average radius of the first load

R CP 1 = (2R + R Ö 2)/2 = R(2 + Ö 2)/2.(5.3.3)

Average radius of the second load

R CP 2 =(0 + R Ö 2)/2 = (RÖ 2)/2.(5.3.4)

The projection of the centrifugal force acting on the first load in the direction of the X axis will be

F 1 = mω 2 SR 1 R SR 1 cos(Ψ /2)sin2Ψ » 2.67mω 2 R cos(Ψ /2)sin2Ψ. (5.3.5)

The projection of the centrifugal force acting on the second load in the direction of the X axis will be

F 2 = mω 2 SR 2 R SR 2 sin(Ψ /2)sin2Ψ » 0.4mω 2 R sin(Ψ /2)sin2Ψ. (5.3.6)

° £ Ψ £36° will be

0.2π

F CP 1 + 2 = (1/0.2 π) ∫ (F 1 + F 2)dΨ » 1.47mω 2 R. (5.3.7)

The average value of the difference between the projections of the centrifugal forces of the first and second loads in the interval 36° £ Ψ £72° will be

0.4π

F CP 1 - 2 = (1/0.2 π) ∫(F 1 - F 2) dΨ » 1.95mω 2 R. (5.3.8)

0.2π

The average value of the sum of the projections of the centrifugal forces of the first and second loads in the interval 72° £ Ψ £90° will be

0.5π

F CP- (1 + 2) = - (1/0.1 π) ∫(F 1 + F 2)dΨ » -3.72mω 2 R. (5.3.9)

0.4π

The average value of the sum of the projections of the centrifugal forces of the first and second loads in the interval 0° £ Ψ £90° will be

F CP = (2F CP 1 + 2 + 2F CP 1 – 2 + F CP- (1 + 2))/5 » 0.62mω 2 R. (5.3.10)

The sum of the projections of centrifugal forces for the third and fourth loads is calculated similarly.

To eliminate the torque, it is necessary to use another pair of rods, but rotating in the opposite direction relative to their own center of mass and relative to the common axis of rotation, then the final traction force will be

F T = 4F CP = 2.48mω 2 R. (5.3.11)

Let m = 0.1kg; ω =2 πf, where f = 10 rev/s; R = 0.25m, then F T ≈ 245N.

§6. Phase force of inertia.

To implement the phase force of inertia as a translational force, we use a two-crank four-link articulated linkage to convert the uniform rotation of the engine into uneven rotation of loads according to a certain mode, optimizing the nature of the movement of loads for effective use inertia forces, and the corresponding choice relative position loads, compensate for the reverse impulse

A four-bar articulated linkage will be double-cranked if the center-to-center distance is AG (Fig. 6.1) will be less than the length of any moving link, and the sum of the center-to-center distance and the length of the largest of the moving links will be less than the sum of the lengths of the other two links.

Rice. 6.1.

The VG link (lever), on which a load of mass m is attached, is a driven crank on a fixed shaft G, and the AB link is a leading one. Link A is the motor shaft. The BV link is a connecting rod. The ratio of the lengths of the connecting rod and the drive crank is selected so that when the load reaches extreme point D there was a right angle between the connecting rod and the drive crank, which ensures maximum efficiency. Then, with uniform rotation of the engine shaft A with the driving crank AB with an angular velocity w, the connecting rod BV transmits movement to the driven crank VG, slowing it down. Thus, the load slows down from point E to point D along the upper semicircle. In this case, the inertial force acts in the direction of movement of the load. Let's consider the movement of the load in the opposite semicircle (Fig. 6.2), where the connecting rod, straightening, accelerates the load.

Rice. 6.2.

In this case, the inertial force acts against the direction of movement of the load, coinciding with the direction of the inertial force in the first semicircle. The integrated propulsion circuit is shown in Figure 6.3.

Rice. 6.3.

The driving cranks AB and A¢ B¢ are rigidly connected in a straight line on the engine shaft, and the driven cranks (levers) rotate independently of each other on a stationary shaft. The longitudinal components of the inertia forces in the direction from point E to point D of the upper and lower loads add up, providing forward motion. There is no reverse impulse, since the weights rotate in the same direction and, on average, are symmetrically oppositely located.

Let us evaluate the effective phase inertia force.

Let AB = BV = r, GV = R.

Let us assume that in the extreme right position the angle Ψ between the radius R and midline DE is equal to 0° (Fig. 6.4) and

r + r – AG = R, (6 .1)

and also in the extreme left position at Ψ =180° (Fig.6.5) angle

Р ABC = 90°. (6 .2)

Then, based on these conditions, it is easy to determine that the assumptions are satisfied for the following values

r = 2R/(2+Ö 2), (6.3)

AG = (3 - 2Ö 2)R. (6 .4)

Now let's determine the angular velocities in the extreme right and left positions. Obviously, in the right position, the angular velocities of the AG and GW coincide and are equal to w.

Rice. 6.4.

In the left position, the angular velocity w of the GW will obviously be equal to

w GW = (180° /225° )w . (6 .5)

The increment in angular velocity ∆w during the time ∆t = 225° /w = 5π/4w will be

∆w = w GW - w = - 0.2w. (6 .6)

Let angular acceleration will be equally slow, then

dω/dt = ∆w /∆t = - 0.16w 2 / π. (6 .7)

Let us use the formula for the phase inertia force (2.8) in scalar form

F f = -m [(dω/dt)R] = 0.16mw 2 R/ π. (6.8)

Rice. 6.5.

The projection of the phase inertia force in the ED direction will be

F fED = 0.16mw 2 RsinΨ/π. (6.9)

Average value of the projection of the phase inertia force for a half-cycle

F CP = 0.16mω 2 R/ π 2) ∫ sinΨdΨ = 0.32mω 2 R/ π 2 . (6.10)

For two loads (Fig. 6.3), the force doubles. To eliminate the torque, it is necessary to apply another pair of weights, but rotating in the opposite direction. Finally, the traction force for four loads will be

F T = 4F CP = 1.28mω 2 R/ π 2. (6.11)

Let m = 0.1kg; ω =2 πf, where f = 10 rev/s; R = 0.5 m, then F T = 25.6 N.

§7. Gyroscope. Coriolis and centrifugal inertial force.

Let's consider oscillatory motion load with mass m along a semicircle (Fig. 7.1) with radius R with linear speed v. The centrifugal force of inertia Fc acting on a load of mass m will be equal to m v 2 /R, directed radially from the center O. The projection of the centrifugal force on the X axis will be equal

F c׀׀ = (m v 2 /R) sin α. (7.1)

The load must move with acceleration w along the circumference, so that the centrifugal force is effective for the translational motion of the system, and since v = wt, then

F c׀׀ = (m w 2 t 2 /R) sin α, (7.2)

where t is time.

Rice. 7.1.

Due to the inertia of the load, a reverse impulse appears at the edges of the semicircle, which prevents the forward movement of the system in the direction of the X axis.

It is known that when exposed to a force that changes the direction of the gyroscope axis, it precesses under the influence of the Coriolis force, and this movement is inertia-free. That is, with the instant application of a force that changes the direction of the rotation axis, the gyroscope instantly begins to precess and just as instantly stops when this force disappears. Instead of a load, we use a gyroscope rotating with angular velocity ω. Now let’s apply a force F perpendicular to the axis of rotation of the gyroscope (Fig. 7.2) and influence the axis so that the holder with the gyroscope performs inertia-free oscillatory motion (precesses) in a certain sector (in the optimal case with final valueα = 180°). An instant stop of the precession of the holder with a gyroscope and its resumption in the opposite direction occurs when the direction of the force F changes to the opposite. Thus, an oscillatory, inertia-free movement of the holder with a gyroscope occurs, which eliminates the reverse impulse that prevents forward movement along the X axis.

Rice. 7.2.

Angular velocity of precession

dα /dt = M / I Z ω, (7.3)

where: M – moment of force; I Z – moment of inertia of the gyroscope; ω – angular velocity of the gyroscope.

Moment of force (assuming ℓ is perpendicular to F)

M = ℓ F, (7.4)

where: ℓ – distance from the point of application of force F to the center of inertia of the gyroscope; F – force applied to the axis of the gyroscope.

Substituting (7.4) into (7.3) we get

dα /dt = ℓ F / I Z ω, (7.5)

On the right side of formula (7.5) the components ℓ, I Z, We consider ω constant, and let the force F, depending on time t, vary according to a piecewise linear law (Fig. 7.3).

Rice. 7.3.

It is known that linear speed is related to angular speed by the following relation

v = R(dα/dt). (7.6)

Differentiating formula (7.6) with respect to time, we obtain acceleration

w = R (d 2 α /dt 2). (7.7)

Substituting formula (7.5) into formula (7.7) we get

w = (R ℓ/IZω ) (dF/dt). (7.8)

Thus, acceleration depends on the rate of change of force F, which makes the centrifugal force effective for the translational motion of the system.

It should be noted that at high angular velocity ω and dα /dt<< ω , возникающий гироскопический момент уравновешивает момент силы F, поэтому движения в направлении воздействия этой силы не происходит .

To compensate for the perpendicular projection of the centrifugal force Fc ┴ we use a second similar gyroscope, which performs an oscillatory movement synchronously in antiphase with the first gyroscope (Fig. 7.4). The projection of the centrifugal force Fc ┴ at the second gyroscope will be directed opposite to the projection at the first. It is obvious that the perpendicular components Fc ┴ will be compensated, and the parallel components Fc׀׀ will be added.

Rice. 7.4.

If the sector of oscillation of the gyroscopes is no more than a semicircle, then the opposite centrifugal force will not arise, reducing the centrifugal force in the direction of the X axis.

To eliminate the torque of the device arising due to the forced rotation of the gyroscope axis, it is necessary to install another pair of the same gyroscopes, the axes of which rotate in the opposite direction. The sectors of oscillatory motion of holders with gyroscopes in pairs, the axes of which gyroscopes rotate in one direction, should be symmetrically directed in one direction with the sectors of holders with gyroscopes, the axes of which gyroscopes rotate in the other direction (Fig. 7.5).

Rice. 7.5.

Let's calculate the average value of the projection Fс׀׀ of the centrifugal force for one gyroscope (Fig. 7.2) on a holder oscillating in the semicircle sector from 0 to π and denote this value by Fп

Fп = (1/ π) ∫ (m w 2 t 2 / R) sin α dα = 2m w 2 t 2 / Rπ. (7.9)

For four gyroscopes on holders, the average value of the translational force Fп for each half-cycle will be:

Fп = 8m w 2 t 2 / Rπ. (7.10)

Let the mass of the holder be much less than the mass of the gyroscope, and the mass of the gyroscope m = 1 kg. Acceleration w = 5 m/s 2 , and the acceleration of the gyroscope is an order of magnitude greater than the acceleration of the system, then we can ignore the small interval of absence of the action of the centrifugal force in the center. Speed ​​rise time t = 1s. Radius (length) of the holder R = 0.5 m. Then, according to formula (7.10), the translational force will be Fп = 8∙ 1∙ 5 2 ∙1 2 /0.5 π ≈ 127N.

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Let's consider a cart with a bracket attached to it, from which a ball is suspended on a thread (Fig. 5.1). While the cart is at rest or moving without acceleration, the thread is vertical and the force of gravity is m g is balanced by the reaction of the thread F r. If we now bring the cart into linear motion with acceleration A = A in , the thread will deviate from the vertical at such an angle that the resulting force m g And F r,. gave the ball an acceleration equal to A in:

m A in =m g + F r. (5.6)

With respect to the reference frame associated with the cart, the ball is at rest, despite the fact that the resultant force m g And F r is different from zero. The lack of acceleration of the ball with respect to this reference frame can be formally explained by the fact that, in addition to the forces m g And F r equal in total to m A in , the ball is acted upon by an inertial force F in = –m A in. In the final case, we get the same equation (5.6).

m a=m g + F r.+ F in =m g + F r. –m A in = 0, (5.7)

Rice. 5.1. Fig.5. 2. Fig 5.3.

The introduction of inertial forces makes it possible to describe the motion of bodies in any (both inertial and non-inertial) reference systems using the same equations of motion.

However, it should be understood that inertial forces cannot be put on a par with forces caused by fundamental interactions, such as gravitational and electromagnetic forces, or elastic and frictional forces. All these forces are caused by the influence on the body from other bodies. Inertial forces are determined by the properties of the reference system in which mechanical phenomena are considered.

The introduction of inertial forces into consideration is not fundamentally necessary. In principle, any movement can always be considered in relation to an inertial frame of reference. However, in practice, it is often the motion of bodies in relation to non-inertial reference systems, for example, in relation to the earth's surface, that is of interest. The use of inertial forces makes it possible to solve the corresponding problem directly in relation to such a reference system, which often turns out to be much simpler than considering motion in an inertial frame.

A characteristic property of inertial forces is their proportionality to the mass of the body. Thanks to this property, inertial forces turn out to be similar to gravitational forces. Let's imagine that we are in a closed cabin remote from all external bodies, which is moving with acceleration g in the direction we will call “up” (Fig. 5.3). Then all bodies located inside the cabin will behave as if they were acted upon by an inertial force F in = –m g. In particular, a spring, to the end of which a body of mass m is suspended, will stretch so that the elastic force balances the inertia force –m g. However, the same phenomena would be observed if the cabin were stationary and located near the surface of the Earth. Without the opportunity to “look” outside the cabin, no experiments carried out inside the cabin would allow us to establish what causes the force –m g– accelerated movement of the cabin or the action of the Earth’s gravitational field. On this basis, they speak of the equivalence of the forces of inertia and gravity (in a uniform gravitational field). This equivalence underlies Einstein's general theory of relativity (GTR).