The force of inertia of a material point. Existence of inertial frames of reference
inertia - the ability to keep one's state unchanged is an intrinsic property of all material bodies.
Inertia force - force arising from acceleration or deceleration of a body (material point) and directed in the opposite direction from acceleration. The force of inertia can be measured, it is applied to "connections" - bodies connected with an accelerating or decelerating body.
It is calculated that the force of inertia is equal to
F in = | m*a|
Thus, the forces acting on material points m 1 and m2(Fig. 14.1), when overclocking the platform, respectively, are equal
F in1 \u003d m 1 * a; F in2 \u003d m 2 * a
Accelerating body (platform with mass t(Fig. 14.1)) does not perceive the force of inertia, otherwise the acceleration of the platform would be impossible at all.
With rotational (curvilinear) motion, the resulting acceleration is usually represented in the form of two components: normal a p and tangent a t(Fig. 14.2).
Therefore, when considering a curvilinear motion, two components of the inertia force can arise: normal and tangential
a = a t + a n ;
With uniform motion along an arc, normal acceleration always occurs, the tangential acceleration is zero, therefore, only the normal component of the inertia force acts, directed along the radius from the center of the arc (Fig. 14.3).
The principle of kinetostatics (d'Alembert's principle)
The principle of kinetostatics is used to simplify the solution of a number of technical problems.
In reality, the forces of inertia are applied to the bodies connected with the accelerating body (to the bonds).
d'Alembert suggested conditionally apply force of inertia to an actively accelerating body. Then the system of forces applied to the material point becomes balanced, and it is possible to use the equations of statics when solving problems of dynamics.
d'Alembert's principle:
A material point under the action of active forces, reactions of bonds and a conditionally applied force of inertia is in equilibrium;
End of work -
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Problems of theoretical mechanics
Theoretical mechanics is the science of the mechanical motion of material solids and their interaction. Mechanical motion is understood as the movement of a body in space and time along
Third axiom
Without violating the mechanical state of the body, you can add or remove a balanced system of forces (the principle of discarding a system of forces equivalent to zero) (Fig. 1.3). P,=P2 P,=P.
Consequence from the second and third axioms
The force acting on a rigid body can be moved along its line of action (Fig. 1.6).
Bonds and Reactions of Bonds
All laws and theorems of statics are valid for a free rigid body. All bodies are divided into free and bound. Free bodies are bodies whose movement is not limited.
Rigid rod
In the diagrams, the rods are depicted with a thick solid line (Fig. 1.9). rod mozhe
fixed hinge
The attachment point cannot be moved. The rod can freely rotate around the hinge axis. The reaction of such a support passes through the hinge axis, but
Planar system of converging forces
The system of forces, the lines of action of which intersect at one point, is called convergent (Fig. 2.1).
Resultant of converging forces
The resultant of two intersecting forces can be determined using a parallelogram or a triangle of forces (4th axiom) (Vis. 2.2).
Equilibrium condition for a flat system of converging forces
When the system of forces is in equilibrium, the resultant must be equal to zero, therefore, in a geometric construction, the end of the last vector must coincide with the beginning of the first. If a
Solving equilibrium problems in a geometric way
It is convenient to use the geometric method if there are three forces in the system. When solving equilibrium problems, the body is considered to be absolutely solid (solidified). The order of solving problems:
Decision
1. The forces arising in the fastening rods are equal in magnitude to the forces with which the rods support the load (5th axiom of statics) (Fig. 2.5a). We determine the possible directions of the reactions of the bond
Projection of Force on the Axis
The projection of the force on the axis is determined by the segment of the axis, cut off by perpendiculars, lowered onto the axis from the beginning and end of the vector (Fig. 3.1).
Forces in an analytical way
The value of the resultant is equal to the vector (geometric) sum of the vectors of the system of forces. We determine the resultant geometrically. We choose a coordinate system, determine the projections of all tasks
Converging forces in analytical form
Based on the fact that the resultant is equal to zero, we get: Condition
Couple of forces, moment of a couple of forces
A pair of forces is a system of two forces that are equal in modulus, parallel and directed in different directions. Consider a system of forces (P; B") forming a pair.
Moment of force about a point
A force that does not pass through the attachment point of the body causes the body to rotate relative to the point, so the effect of such a force on the body is estimated as a moment. Moment of force rel.
Poinsot's theorem on parallel transfer of forces
A force can be transferred parallel to its line of action by adding a pair of forces with a moment equal to the product of the modulus of the force and the distance over which the force has been transferred.
located forces
The lines of action of an arbitrary system of forces do not intersect at one point, therefore, to assess the state of the body, such a system should be simplified. To do this, all the forces of the system are transferred to one arbitrarily
Influence of the reference point
The reference point is chosen arbitrarily. When you change the position of the reduction point, the value of the main vector will not change. The value of the main moment when the reduction point is moved will change,
Flat force system
1. At equilibrium, the main vector of the system is equal to zero. The analytical definition of the main vector leads to the conclusion:
Types of loads
According to the method of application, loads are divided into concentrated and distributed. If in reality the transfer of the load occurs on a negligible area (at a point), the load is called concentrated
Moment of force about the axis
The moment of force about the axis is equal to the moment of the projection of the force onto a plane perpendicular to the axis, relative to the point of intersection of the axis with the plane (Fig. 7.1 a). MO
Vector in space
In space, the force vector is projected onto three mutually perpendicular coordinate axes. The projections of the vector form the edges of a rectangular parallelepiped, the force vector coincides with the diagonal (Fig. 7.2
Spatial convergent system of forces
A spatial converging system of forces is a system of forces that do not lie in the same plane, the lines of action of which intersect at one point. The resultant spatial system si
Bringing an arbitrary spatial system of forces to the center O
A spatial system of forces is given (Fig. 7.5a). Let us bring it to the center O. Forces must be moved in parallel, and a system of pairs of forces is formed. The moment of each of these pairs is
Center of gravity of homogeneous flat bodies
(flat figures) Very often it is necessary to determine the center of gravity of various flat bodies and geometric flat figures of complex shape. For flat bodies, we can write: V =
Determining the coordinates of the center of gravity of flat figures
Note. The center of gravity of a symmetrical figure is on the axis of symmetry. The center of gravity of the rod is at the middle of the height. The positions of the centers of gravity of simple geometric shapes can
Point kinematics
Have an idea about space, time, trajectory, path, speed and acceleration. Know how to set the movement of a point (natural and coordinate). Know the notation
Distance traveled
The path is measured along the path in the direction of travel. Designation - S, units of measurement - meters. Point motion equation: Equation defining
Travel speed
The vector value that characterizes at the moment the speed and direction of movement along the trajectory is called speed. Velocity is a vector directed along k
point acceleration
The vector quantity characterizing the rate of change of speed in magnitude and direction is called the acceleration of a point. Point speed when moving from point M1
Uniform movement
Uniform motion is motion at a constant speed: v = const. For rectilinear uniform motion (Fig. 10.1 a)
Equal-variable motion
Equal-variable motion is motion with constant tangential acceleration: at = const. For rectilinear uniform motion
translational movement
Translational is such a movement of a rigid body in which any straight line on the body during movement remains parallel to its initial position (Fig. 11.1, 11.2). At
rotational movement
During rotational motion, all points of the body describe circles around a common fixed axis. The fixed axis around which all points of the body rotate is called the axis of rotation.
Particular cases of rotational motion
Uniform rotation (the angular velocity is constant): ω = const The equation (law) of uniform rotation in this case has the form:
Speeds and accelerations of points of a rotating body
The body rotates around the point O. Let's determine the motion parameters of the point Alocated at a distance RA from the axis of rotation (Fig. 11.6, 11.7). Way
Decision
1. Section 1 - uneven accelerated movement, ω \u003d φ '; ε = ω’ 2. Section 2 - the speed is constant - the movement is uniform, . ω = const 3.
Basic definitions
A complex movement is a movement that can be decomposed into several simple ones. Simple movements are translational and rotational. To consider the complex movement of points
Plane-parallel motion of a rigid body
Plane-parallel, or flat, is such a motion of a rigid body in which all points of the body move parallel to some fixed one in the reference frame under consideration
translational and rotational
Plane-parallel motion is decomposed into two motions: translational along with a certain pole and rotational with respect to this pole. Decomposition is used to determine
Center of speeds
The speed of any point of the body can be determined using the instantaneous center of velocities. In this case, a complex movement is represented as a chain of rotations around different centers. Task
Axioms of dynamics
The laws of dynamics summarize the results of numerous experiments and observations. The laws of dynamics, which are usually considered as axioms, were formulated by Newton, but the first and fourth laws were also
The concept of friction. Types of friction
Friction is the resistance that occurs when one rough body moves over the surface of another. When bodies slide, sliding friction arises; when rolling, rolling friction occurs. The nature of resistance
rolling friction
The rolling resistance is related to the mutual deformation of the ground and the wheel and is much less than the sliding friction. Usually the soil is considered softer than the wheel, then the soil is mainly deformed, and
Free and non-free points
A material point, the movement of which in space is not limited by any constraints, is called free. Problems are solved using the basic law of dynamics. Material then
Decision
Active forces: driving force, friction force, gravity. Reaction in the support R. We apply the force of inertia in the opposite direction from the acceleration. According to the d'Alembert principle, the system of forces acting on the platform
The work of the resultant force
Under the action of a system of forces, a point of mass m moves from position M1 to position M 2 (Fig. 15.7). In the case of movement under the action of a system of forces,
Power
To characterize the performance and speed of work, the concept of power is introduced. Power is the work done per unit of time:
Rotating power
Rice. 16.2 The body moves along an arc of radius from point M1 to point M2 M1M2 = φr Work of force
Efficiency
Each machine and mechanism, doing work, spends part of the energy to overcome harmful resistances. Thus, the machine (mechanism), in addition to useful work, also performs an additional
Theorem on the change in momentum
The momentum of a material point is a vector quantity equal to the product of the mass of the point and its velocity mv. The momentum vector coincides with
Kinetic energy change theorem
Energy is the ability of a body to perform mechanical work. There are two forms of mechanical energy: potential energy, or positional energy, and kinetic energy,
Fundamentals of the dynamics of the system of material points
A set of material points interconnected by interaction forces is called a mechanical system. Any material body in mechanics is considered as a mechanical
The basic equation of the dynamics of a rotating body
Let a rigid body rotate around the Oz axis under the action of external forces with an angular velocity
Voltage
The section method makes it possible to determine the magnitude of the internal force factor in the section, but does not make it possible to establish the law of distribution of internal forces over the section. To assess the strength of n
Internal force factors, stresses. Plotting
Have an idea about longitudinal forces, about normal stresses in cross sections. Know the rules for constructing diagrams of longitudinal forces and normal stresses, the law of distribution
Longitudinal forces
Consider a beam loaded with external forces along the axis. The beam is fixed in the wall (fixing "embedding") (Fig. 20.2a). We divide the beam into sections of loading. Load area with
Geometric characteristics of flat sections
Have an idea about the physical meaning and procedure for determining the axial, centrifugal and polar moments of inertia, about the main central axes and the main central moments of inertia.
Static moment of sectional area
Consider an arbitrary section (Fig. 25.1). If we divide the section into infinitely small areas dA and multiply each area by the distance to the coordinate axis and integrate the obtained
centrifugal moment of inertia
The centrifugal moment of inertia of a section is the sum of the products of elementary areas taken by the total area by both coordinates:
Axial moments of inertia
The axial moment of inertia of the section with respect to some yard lying in the same plane is the sum of the products of elementary areas per square of their distance taken over the entire area
Polar moment of inertia of the section
The polar moment of inertia of a section relative to a certain point (pole) is the sum of the products of elementary areas taken over the entire area and the square of their distance to this point:
Moments of inertia of the simplest sections
Axial moments of inertia of a rectangle (Fig. 25.2) Let's imagine directly
Polar moment of inertia of a circle
For a circle, the polar moment of inertia is first calculated, then the axial ones. Imagine a circle as a set of infinitely thin rings (Fig. 25.3).
Torsional deformations
The torsion of a round beam occurs when it is loaded by pairs of forces with moments in planes perpendicular to the longitudinal axis. In this case, the beam generatrix is bent and turned through an angle γ,
Hypotheses in torsion
1. The hypothesis of flat sections is fulfilled: the beam cross section, which is flat and perpendicular to the longitudinal axis, remains flat and perpendicular to the longitudinal axis after deformation.
Internal force factors in torsion
Torsion is called loading, in which only one internal force factor arises in the cross section of the beam - torque. External loads are also two pro
Torque Plots
Torques can vary along the axis of the beam. After determining the values of the moments along the sections, we build a plot of torques along the axis of the bar.
Torsional stresses
We draw a grid of longitudinal and transverse lines on the surface of the beam and consider the pattern formed on the surface after Fig. 27.1a deformation (Fig. 27.1a). Pop
Maximum torsional stresses
It can be seen from the formula for determining the stresses and the diagram of the distribution of shear stresses during torsion that the maximum stresses occur on the surface. Determine the maximum voltage
Types of strength calculations
There are two types of strength calculation 1. Design calculation - the diameter of the beam (shaft) in the dangerous section is determined:
Stiffness calculation
When calculating the stiffness, the deformation is determined and compared with the allowable one. Consider the deformation of a round bar under the action of an external pair of forces with a moment t (Fig. 27.4).
Basic definitions
A bend is a type of loading in which an internal force factor arises in the cross section of the beam - a bending moment. Bar working on
Internal force factors in bending
Example 1. Let's consider a beam, which is acted upon by a pair of forces with a moment t and an external force F (Fig. 29.3a). To determine the internal force factors, we use the method with
Bending moments
The transverse force in the section is considered positive if it tends to turn the
Differential dependencies for direct transverse bending
The construction of diagrams of shear forces and bending moments is greatly simplified when using differential relationships between the bending moment, shear force and uniform intensity.
Section method The resulting expression can be generalized
The transverse force in the section under consideration is equal to the algebraic sum of all forces acting on the beam up to the section under consideration: Q = ΣFi Since we are talking about
Voltage
Consider the bending of a beam pinched on the right and loaded with a concentrated force F (Fig. 33.1).
Stress state at a point
The stress state at a point is characterized by normal and tangential stresses arising on all areas (sections) passing through the given point. It is usually sufficient to define
The concept of a complex deformed state
The set of strains that occur in different directions and in different planes passing through a point determine the deformed state at that point. Complex deformi
Calculation of a round bar for bending with torsion
In the case of calculating a round beam under the action of bending and torsion (Fig. 34.3), it is necessary to take into account normal and shear stresses, since the maximum stress values in both cases occur
The concept of stable and unstable equilibrium
Relatively short and massive rods rely on compression, because. they fail as a result of destruction or residual deformations. Long rods of small cross-section under the action
Sustainability calculation
The stability calculation consists in determining the permissible compressive force and, in comparison with it, the acting force:
Calculation by the Euler formula
The problem of determining the critical force was mathematically solved by L. Euler in 1744. For a rod hinged on both sides (Fig. 36.2), the Euler formula has the form
Critical stresses
Critical stress is the compressive stress corresponding to the critical force. The stress from the compressive force is determined by the formula
Limits of applicability of the Euler formula
The Euler formula is valid only within the limits of elastic deformations. Thus, the critical stress must be less than the elastic limit of the material. Prev
AT classical mechanics ideas about forces and their properties are based on Newton's laws and are inextricably linked with the concept inertial frame of reference.
Indeed, the physical quantity called force is introduced into consideration by Newton's second law, while the law itself is formulated only for inertial frames of reference. Accordingly, the concept of force initially turns out to be defined only for such frames of reference.
Newton's second law equation relating acceleration and mass material point with the force acting on it, is written in the form
It directly follows from the equation that only forces are the cause of acceleration of bodies, and vice versa: the action of uncompensated forces on a body necessarily causes its acceleration.
Newton's third law complements and develops what was said about forces in the second law.
force is a measure of the mechanical action on a given material body of other bodies
in accordance with Newton's third law, forces can only exist in pairs, and the nature of the forces in each such pair is the same.
any force acting on a body has a source of origin in the form of another body. In other words, forces are necessarily the result interactions tel.
No other forces in mechanics are brought into consideration or used. The possibility of the existence of forces that have arisen independently, without interacting bodies, is not allowed by mechanics.
Although the names of the Euler and d'Alembert forces of inertia contain the word force, these physical quantities are not forces in the sense accepted in mechanics.
34. The concept of plane-parallel motion of a rigid body
The motion of a rigid body is called plane-parallel if all points of the body move in planes parallel to some fixed plane (the main plane). Let some body V make a plane motion, π - the main plane. From definitions plane-parallel motion and the properties of an absolutely rigid body, it follows that any segment of the straight line AB, perpendicular to the plane π, will perform translational motion. That is, the trajectories, speeds and accelerations of all points of the segment AB will be the same. Thus, the movement of each point of the section s parallel to the plane π determines the movement of all points of the body V lying on the segment perpendicular to the section at this point. Examples of plane-parallel motion are: wheel rolling along a straight segment, since all its points move in planes parallel to the plane perpendicular to the wheel axis; a special case of such a movement is rotation of a rigid body around a fixed axis, in fact, all points of the rotating body move in planes parallel to some fixed plane perpendicular to the axis of rotation.
35. Forces of inertia in the rectilinear and curvilinear motion of a material point
The force with which a point resists a change in motion is called the force of inertia of a material point. The force of inertia is directed opposite to the acceleration of the point and is equal to the mass times the acceleration.
In a straight line the direction of acceleration coincides with the trajectory. The force of inertia is directed in the direction opposite to acceleration, and its numerical value is determined by the formula:
With accelerated movement, the directions of acceleration and speed coincide and the force of inertia is directed in the direction opposite to the movement. In slow motion, when the acceleration is directed in the direction opposite to the speed, the force of inertia acts in the direction of motion.
Atcurvilinear and unevenmovement acceleration can be decomposed into normal an and tangent at components. Similarly, the force of inertia of a point also consists of two components: normal and tangential.
Normal the component of the inertial force is equal to the product of the mass of the point and the normal acceleration and is directed opposite to this acceleration:
Tangent the component of the inertial force is equal to the product of the mass of the point and the tangential acceleration and is directed opposite to this acceleration:
Obviously, the total force of inertia of the point M is equal to the geometric sum of the normal and tangent components, i.e.
Given that the tangent and normal components are mutually perpendicular, the total inertia force.
Having established that the individual points in Newtonian absolute space are not a physical reality, we must now ask ourselves what remains within
this concept at all? The following remains: the resistance of all bodies to acceleration must be interpreted in the Newtonian sense as an action of absolute space. The locomotive that sets the train in motion overcomes the resistance of inertia. A projectile that takes down a wall draws its destructive force from inertia. The action of inertia is manifested whenever accelerations take place, and the latter are nothing more than changes in velocity in absolute space (we can use the latter expression, since the change in velocity has the same magnitude in all inertial frames). Thus, frames of reference, which themselves move with acceleration relative to inertial frames, 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 take on a more complex form. Even the trajectory of a free body turns out to be no longer uniform and rectilinear in an accelerated system (see Chap. 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. The body, which is not affected by real forces, is still subject to the action of these inertial forces, so its movement in the general case turns out to be uneven and non-rectilinear. For example, a car that starts moving or slows down is such an accelerated system. Everyone knows the push of a moving or stopping train; it is nothing but 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 the law of motion in the Newtonian form is satisfied in the accelerated system, namely
except that now the force must be set to K, which is equal to
where K is the real force, and is the apparent force, or the force of inertia.
So, this force acts on a free body. Its action can be illustrated by the following reasoning: we know that gravity on Earth - the force of gravity - is determined by the formula G = mg, where is a constant acceleration due to gravity. The force of inertia acts in this case like gravity; the minus sign means that the force of inertia is directed opposite to the acceleration of the frame of reference, which is used as a basis. The magnitude of the apparent gravitational acceleration y coincides with the acceleration of the frame of reference. Thus, the motion of a free body in the frame is simply motion of the type that we know as the fall or motion of a thrown body.
This relationship between inertial forces in accelerated systems and the force of gravity here still seems somewhat artificial. 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 of relativity.
FORCE OF INERTIA
FORCE OF INERTIA
A vector quantity numerically equal to the product of the mass m of a material point and its w and directed oppositely to the acceleration. At curvilinear movement of S. and. can be decomposed into a tangent or tangential component Jt, directed oppositely to the tangent. acceleration wt , and on the normal component Jn directed along the normal to the trajectory from the center of curvature; numerically Jt=mwt, Jn=mv2/r, where v - points, r - radius of curvature of the trajectory. When studying motion in relation to the inertial frame of reference, S. and. introduced in order to have a formal opportunity to compose dynamics equations in the form of simpler statics equations (see). The concept of S. and. is also introduced in the study of relative motion. In this case, the addition of forces of interaction with other bodies acting on a material point S. and. - portable Jper and Coriolis forces Jcor - allows you to compose equations of motion of this point in a moving (non-inertial) frame of reference in the same way as in inertial.
Physical Encyclopedic Dictionary. - M.: Soviet Encyclopedia. . 1983 .
FORCE OF INERTIA
A vector quantity numerically equal to the product of the mass t material point on its acceleration w and directed opposite to the acceleration. At curvilinear movement of S. and. can be decomposed into a tangent, or tangential, component directed oppositely to the tangent. acceleration, and on the normal, or centrifugal, component, directed along Ch. trajectory normals from the center of curvature; numerically , , where v- the speed of the point is the radius of curvature of the trajectory. When studying movement in relation to inertial frame of reference S. i. introduced in order to have a formal opportunity to compose dynamics equations in the form of simpler statics equations (see. D "Alamber principle, Kinetostatics).
The concept of S. and. also introduced in the study relative movement. In this case, by adding the transfer force J nep to the forces of interaction with other bodies acting on the material point and Coriolis force inertia, Targ.
Physical encyclopedia. In 5 volumes. - M.: Soviet Encyclopedia. Editor-in-Chief A. M. Prokhorov. 1988 .
See what the "POWER OF INERTIA" is in other dictionaries:
- (also inertial force) a term widely used in various meanings in the exact sciences, and also, as a metaphor, in philosophy, history, journalism and fiction. In the exact sciences, the force of inertia is usually a concept ... Wikipedia
Modern Encyclopedia
A vector quantity numerically equal to the product of the mass m of a material point and the module of its acceleration? and directed opposite to the acceleration... Big Encyclopedic Dictionary
inertia force- A vector quantity, the module of which is equal to the product of the mass of a material point and the module of its acceleration and directed opposite to this acceleration. [Collection of recommended terms. Issue 102. Theoretical Mechanics. USSR Academy of Sciences. The committee… … Technical Translator's Handbook
inertia force- INERTIA FORCE, a vector quantity numerically equal to the product of the mass m of a material point and its acceleration u and directed oppositely to the acceleration. It arises due to the non-inertiality of the reference system (rotation or rectilinear motion with ... ... Illustrated Encyclopedic Dictionary
inertia force- inercijos jėga statusas T sritis Standartizacija ir metrologija apibrėžtis Vektorinis dydis, lygus materialiojo taško arba kūno masės ir pagreičio sandaugai; kryptis priešinga pagreičiui. atitikmenys: engl. inertia force vok. Tragheitskraft, f;… … Penkiakalbis aiskinamasis metrologijos terminų žodynas
A vector quantity numerically equal to the product of the mass m of a material point and the modulus of its acceleration w and directed oppositely to the acceleration. * * * FORCE OF INERTIA FORCE OF INERTIA, a vector quantity, numerically equal to the product of the mass m of the material ... ... encyclopedic Dictionary
inertia force- inercijos jėga statusas T sritis automatika atitikmenys: angl. inertial force vok. Tragheitskraft, f rus. inertial force, fpranc. force d inertie, f … Automatikos terminų žodynas
inertia force- inercijos jėga statusas T sritis fizika atitikmenys: angl. inertial force vok. Tragheitskraft, f rus. inertial force, fpranc. force d'inertie, f … Fizikos terminų žodynas
inertia force- a value numerically equal to the product of the body mass and its acceleration and directed opposite to the acceleration; See also: force force of friction force of light force of drag force of internal friction ... Encyclopedic Dictionary of Metallurgy
Forces of inertia and the basic law of mechanics
Bernikov Vasily Ruslanovich,
engineer.
Foreword
Internal forces in some cases are the cause of the appearance of external forces applied to the system , , , . Forces of inertia 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, the possibility of using inertial forces as a translational force in creating structures did not lead to a positive result. Only some well-known designs with a low efficiency of using inertia forces can be noted: Tolchin's inertsoid, Frolov's vortex liquid propulsor, Thornson's propulsor. The slow development of inertial propulsion is explained by the lack of a fundamental theoretical substantiation of the observed effect. On the basis of the usual classical concepts of physical mechanics, in this work, a theoretical basis for the use of inertia forces as a translational force has been created.
§one. Basic law of mechanics and its consequences.
Let's consider the laws of transformation of forces and accelerations in different frames of reference. Let us choose an arbitrarily immobile inertial frame of reference and agree that the motion relative to it be considered absolute. In such a reference frame, the basic equation of motion of a material point is the equation expressing Newton's second law.
m w abs = F, (1.1)
where F- the force of interaction of bodies.
A body at rest in a moving frame of reference is dragged by the latter in its motion relative to the fixed frame of reference. This movement is called portable. The motion of a body relative to the reference system is called relative. The absolute motion of a body is made up of its relative and figurative motions. In non-inertial frames of reference (frames of reference moving with acceleration), the law of transformation of accelerations for translational motion has the following form
w abs = w rel +w per. (1.2)
Taking into account (1.1) for forces, we write the equation of relative motion for a material point in a frame of reference moving with translational acceleration
mw rel = F - mw lane, (1.3)
where mw per is the translational force of inertia, which arises not due to the interaction of bodies, but due to the accelerated motion of the reference frame. The movement of bodies under the action of inertial forces is similar to the 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 momentum or the internal translational momentum. Forces of inertia are always external [ 2, p.359] in relation to any moving system of material bodies.
Let us now assume that the frame of reference moves quite arbitrarily relative to the fixed frame of reference. This movement can be divided into two: translational movement with a speed v o, equal to the speed of movement of the origin, and rotational movement around the instantaneous axis passing through this origin. We denote the angular velocity of this rotation w, and the distance from the origin of coordinates of the moving reference system to the moving point in it through r. In addition, the moving point has a speed relative to the moving frame of reference v rel. Then for the absolute acceleration [2, p.362] we know the relation
w abs = w rel - 2[ v rel w] + (d v o /dt) - w 2 r ^ + [ (d w/ dt) r] ,. (1.4)
where r ^ - component of the radius-vector r, perpendicular to the instantaneous axis of rotation. We transfer the relative acceleration to the left side, and the absolute acceleration to the right side and multiply everything by the mass of the body, we get the basic equation of the forces of relative motion [ 2, p.364] of a material point in an arbitrarily moving frame of reference
mw rel = mw abs + 2m[ v rel w] - m(d v o /dt) + mw 2 r ^ – m[ (d w/ dt) r] . (1.5)
Or respectively
mw rel = F + F to + F n+ F c + Fφ, (1.6)
where: F- force of interaction of bodies; F k is the Coriolis force of inertia; F p is the translational force of inertia; F c - centrifugal force of inertia; F f is the phase force of inertia.
The direction of the force of interaction of bodies F coincides with the direction of acceleration of the body. Coriolis force of inertia F k is directed according to the vector product of the radial and angular velocity, that is, perpendicular to both vectors. Translational inertia force 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 force of inertia Fφ is directed opposite to the vector product of the angular acceleration and the 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 in order to determine the trajectory of the body relative to any frame of reference.
In addition to the forces of inertia and the interaction of bodies, there are forces of variable mass, which are the result of the action of inertia forces. Consider Newton's second law in differential form [2, p.77]
d P/dt = ∑ F, (1.7)
where: P is the momentum of the system of bodies; ∑ F is 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) is the mass of the system of bodies; v(t) is the speed of the system of bodies.
Since the speed is the time derivative of the coordinates of the system, then
v(t) = d r(t)/dt, (1.9)
where r is the radius vector.
In the future, we will mean the dependence on time: mass, velocity and radius vector. Substituting (1.9) and (1.8) into (1.7) we obtain
d(m (d r/dt))/dt = ∑ F. (1.10)
We introduce the mass m under the sign of the differential [ 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 for differentiating products
m(d2 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)
We present similar terms and write equation (1.11) in the following form
m(d2 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 variable mass force, that is
F pm = - (dm/dt)(d r/dt). (1.13)
Thus, one more 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 mass change, 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 at a 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 of internal forces (particle separation). Since equation (1.12) was derived under the assumption of a change in the momentum of the system of bodies under the influence of internal forces that generate external ones, by an exact mathematical method, therefore, when it was derived in expression (1.11), two more forces appeared that do not participate in the change in the momentum of the system of bodies, since they cancel when similar terms are reduced. Let us rewrite equation (1.11), taking into account equation (1.13), without canceling like terms, as follows
m(d2 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(d2 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)
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 q also does not participate in the change in the momentum of the system of bodies, so we 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 force of pressure exerted by a substance in a liquid or gaseous state on 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 F q coincides with the variable mass force F pm and their distinction is made only to determine the nature of the action in various conditions. Thus, equation (1.15) completely describes the state of matter. That is, considering equation (1.15), we can conclude that the 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.
The translational accelerated motion of a body occurs under the action of a force according to Newton's second law. That is, a change in the magnitude of the body's velocity occurs in the presence of acceleration and the force that caused this acceleration.
The use of the centrifugal force of inertia for translational motion is possible only with an increase in the linear velocity of the sources of these forces, since with an accelerated movement of the system, the inertia forces of the sources in the direction of increasing the speed of the system decrease until they completely disappear. In addition, the field of inertial forces must be non-uniform and have a maximum value in the part of the system in the direction of translational motion.
Consider the motion of a body (Fig. 2.1) with mass m along a circle with radius R.
Rice. 2.1.
Centrifugal force F c, with which the body presses on the circle, is determined by the formula
F c \u003d m ω 2 R. (2.1)
Using the well-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 \u003d m v 2 / R. (2.2)
The centrifugal force acts in the direction of the radius R. Now let's instantly break the circle along which the body moves. Experience shows that the body will fly tangentially in the direction of the linear velocity v rather than 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 a radius R, and the semicircle moves with acceleration w P perpendicular to the diameter.
Rice. 2.2.
With a uniform motion of the body (linear velocity 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 zero. If the body moves with a 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, it presses on the semicircle. To do this, the time spent by the body on the tangential path to the intersection with the dashed line parallel to the diameter and drawn through point B (Fig. 2.2) must be less than or equal to the time that the semicircle will spend 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 traveled by the semicircle S AB will be
S AB \u003d w P t 2 / 2. (2.4)
We divide equation (2.3) by (2.4) and obtain
S AC / S AB \u003d 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 acceleration of the body in a circle element on 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 the centrifugal force in action. That is, the centrifugal force acts as a translational driving force only in the presence of a positive acceleration that changes the value of the linear velocity of the body in the system
Similarly, the ratio for the second quarter of the semicircle is obtained (Fig. 2.3).
Rice. 2.3.
Only the path traveled by the body along the tangent will start from a point on the semicircle moving with acceleration until it intersects with a dashed line parallel to the diameter and passing through point A of the initial position of the semicircle. The angle in this case is determined by the interval π/2 ³ Ψ ³ 0.
For a system in which the 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 the body rotates with an angular velocity ω and simultaneously approaches the center of the circle with a speed v, then there is a Coriolis force
F k = 2m [ v ω]. (2.6)
A typical trajectory element is shown in Figure 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 during the accelerated motion of the system, the body moving with positive linear acceleration will keep up with the acceleration of the system and , respectively, to move along a curved path, and not along a tangent straight 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 lower one in the interval -π/2 £ Ψ £ π/2, and in the second half from the lower point to the center of the circle π/2 ³ Ψ ³ 0. Similarly, for rotation of the body and its simultaneous removal (Fig. 2.5) from the center, the Coriolis force acts as a translational force 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 0 £ Ψ £ π/2, and in the second half from the bottom point to the end point π/2 ³ Ψ ³ -π/2.
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 is the acceleration of the body.
Rice. 2.6.
With a positive acceleration of the body, it acts against the movement, and with a negative acceleration (deceleration), it acts in the direction of the body's movement. 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, obviously, must be greater than the module of the system's acceleration, caused by the translational force of the body's inertia. That is, the translational force of inertia acts as a driving force in the presence of positive or negative acceleration.
Phase force of inertia F f (the force of inertia caused by uneven rotation) is determined by the formula
F f = -m [(d ω /dt) R]. (2.8)
Let the radius R perpendicular to the angular velocity vector ω , then formula (2.8) in scalar form takes the form
F f \u003d -m (dω / dt) R. (2.9)
With a positive angular acceleration of the body (Fig. 1.7), it acts against the movement, and with a negative angular acceleration (deceleration), it acts in the direction of the body's movement.
Rice. 2.7.
Using the well-known relation ω = v /R, where v is the linear velocity of the body perpendicular to the radius R, we write the formula (2.9) in the following form
F f \u003d -m (dv / dt). (2.10)
Since dv/dt = w , where w is the linear acceleration of the body, 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 the 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 f II , since it is directed in the upper and lower parts of the semicircle in one direction, coinciding with the direction w II .
F fII \u003d -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 force of inertia acts as a driving force in the presence of positive or negative linear acceleration.
So, four elements of the translational inertia force are distinguished: centrifugal, Coriolis, translational, phase. By connecting individual elements in a certain way, it is possible to create systems of translational inertia driving force.
Consider the force of a variable mass defined by the formula
F pm = - (dm/dt)(d r/dt). (2.14)
Since the rate of detachment (attachment) of particles relative to the system of bodies is equal to
u=d r/dt, (2.15)
then equation (2.14) can be written as
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 during the change in its speed from zero to u or the value produced by the acceding particle during the change in its speed from u down to zero. Thus, the variable mass force acts at the moment of acceleration or deceleration of particles, that is, it is a translational inertia force, but calculated by other parameters. In view of the above, it becomes necessary 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 acceleration of the system
d 2 r / dt 2 \u003d 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, it is possible to determine the final speed of the system v by the ratio of the initial m O and final mass m
v = -u O ∫ dm /m = u O ln(m O /m). (2.20)
m O / m \u003d e v / uo. (2.21)
Equation (2.21) is the Tsiolkovsky equation.
§3. The contour of the circulating medium of the centrifugal force of inertia.
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 centrifugal force acting on a point element of the flow 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 along the radius from the center, stretching the ring. The 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, the perpendicular projection of the force will be maximum, gradually decreasing to 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 oppositely directed.
Rice. 3.2.
We write the elementary function of the centrifugal force acting on a point segment with a 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 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 we substitute (3.4) into (3.1), then
∆ F= ρ π r 2 ω 2 R 2 ∆ Ψ.
Centrifugal force acting in a perpendicular direction (Fig. 2)
∆ F┴ = ∆ Fcos((π/2)-Ψ).
It is known that cos((π/2)- Ψ)=sin Ψ, then
∆ F┴ = ∆ F sin Ψ.
Substitute the value for ∆ F we get
∆ F┴ = ρ π r 2 ω 2 R 2 sin Ψ ∆ Ψ.
Find the total centrifugal force acting in the perpendicular direction in the range from 0 to Ψ
F ┴ = ∫ ρ π r 2 ω 2 R 2 sin ΨdΨ.
We 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 greater than the acceleration of the system w c, i.e.
In this case, according to formula (2.5), we obtain
Calculate the angle of action of the forces of inertia in radians
Ψ ≈ 0.467 π,
which corresponds to an angle of 84 degrees.
Thus, the angular interval of action of inertial 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 interval of absence of active inertia forces in the entire circuit is about 6.7% (actually, the acceleration of the circulating medium is much greater than the acceleration of the system, so the interval of absence of active inertia forces will be less than 1% and can be ignored). To determine the total centrifugal force, in these intervals of angles, 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 ┴ \u003d 1.8 ρ π r 2 ω 2 R 2.
It is known that the angular velocity
F ┴ \u003d 1.8 ρ π r 2 v 2.
Since the circulating medium must move with acceleration in order for the inertial force to act, therefore, we express the linear velocity in terms of acceleration, assuming the initial velocity to be zero
F ┴ \u003d 1.8 ρ π r 2 (w t) 2. (3.8)
The average value for the duration of the positive acceleration, which we consider constant, will be
F ┴CP \u003d ((1.8ρ π r 2 w 2) / t) ∫t 2 dt.
After calculations, we get
F ┴CP \u003d 0.6ρ π r 2 w 2 t 2. (3.9).
Thus, the contour of the circulating medium was identified, from which it is possible to make a closed circuit and sum up their centrifugal forces.
Let us compose a closed circuit of four contours of different sections (Fig. 3.3): two upper contours with a radius R. with a section S and two lower contours with a radius R 1 with a section S 1, neglecting the edge effects when the circulating medium passes from one section to another. Let S< S 1 и радиус
R1< 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, we write down the obvious relation for velocities and accelerations
v/v 1 = w / w 1 . (3.11)
Let us find the acceleration of the medium of the lower contour, using equations (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 circuit, taking into account equation (3.12) and after calculations we get
F ┴СР1 = 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 ┴СР (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 value (r 2 / r 1 2).
That is, for 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-planes 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. Calculate the total F C centrifugal force
F C \u003d 2 F ┴SR - 2F ┴SR1 \u003d 1.2ρ π r 2 w 2 t 2 (1- (r 2 / r 1 2)) (3.14)
As you can see, the total centrifugal force depends on the density of the flow, the sections of the opposite contours and the acceleration of the flow. 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 in a circle, the centrifugal force will not cause a translational accelerated motion of the system.
Thus, the basic contour of the circulating medium was identified, the possibility of using the contours of the circulating medium of different sections to sum the centrifugal force in a certain direction and change the total momentum of a closed system of bodies under the action of external inertial forces caused by internal forces was shown.
Let r = 0.025m; r 1 \u003d 0.05 m; ρ \u003d 1000 kg / m 3; w \u003d 5m / s 2, t = 1s, then during the positive acceleration the average value total centrifugal force F C.≈ 44N.
§4. The contour of the circulating medium of the Coriolis force of inertia.
It is known that the Coriolis force of inertia arises when a body of mass m rotates around a circle and simultaneously moves it radially, and it is perpendicular to the angular velocity ω and speed of radial movement v. Direction of the Coriolis Force F coincides with the direction of the vector product in the formula F= 2m[ vw].
Rice. 4.1.
Figure 4.1 shows the direction of the Coriolis force when the body rotates in a circle counterclockwise and moves it radially to the center of the circle in the first half-cycle. and Fig.4.2 shows the direction of the Coriolis force when the body rotates around the circle also counterclockwise and moves it radially from the center of the circle in 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 the movement of the body for the period.
Rice. 4.3.
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. The components of the Coriolis force of the left and right curves F| | AC parallel to the straight line compensate each other, since they are the same, oppositely directed and symmetrical about the X axis. The symmetrical components of the Coriolis force of the left and right F^ curves perpendicular to the straight line AC add up, since they are directed in one direction.
Let us calculate the value of the Coriolis force acting along the X axis on the left half of the trajectory. Since the compilation of the trajectory equation is a difficult task, we are looking for a solution for finding the Coriolis force using an approximate method. Let v be the velocity of the fluid constant along the entire trajectory. The radial speed v p and the linear rotation speed v l, according to the parallelogram theorem of velocities, we express (Fig. 3) through the speed v and the angle α
v p \u003d v cosα, v l \u003d v sinα.
The trajectory of movement (Fig. 4.3) is built taking into account the fact that at point B the radial velocity v p is equal to zero, and the linear velocity v l is equal to v. At the center of the 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 of the trajectory in the center of the circle is perpendicular to the tangent of the trajectory at the beginning (point B). The radius decreases monotonically from Ro to zero. The angle α changes from 90° at point B to 0° at the center of the circle. Then, from graphic constructions, we choose the length of the trajectory 1/4 of the circumference of the circle with radius R 0 . Now you can calculate the mass of a 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 inner radius of the pipe r
m = ρπ 2 r 2 R 0 /2, (4.1)
where ρ is the density of the liquid.
The module of the projection of the Coriolis force at each point of the trajectory on the X axis is found by the formula
F^ = 2m v р ср ω ср cos b , (4.2)
where v p cf is the average value of the radial velocity; ω cf is the average value of the angular velocity; b is the angle between the Coriolis force F and the X axis (-90° £ b £ 90° ).
For technical calculations, it is possible not to take into account the interval of absence of the action of inertial forces, since the acceleration of the circulating medium is much greater than the acceleration of the system. That is, we choose the angle interval between the Coriolis force F and the X axis (-90° £ b £ 90°). The angle α changes from 90° at point B to 0° at the center of the circle, then the average value of the radial velocity
v p cf = 1 / (0 - π/2) ∫ v cos α dα = 2 v / π. (4.3)
The average value of the angular velocity will be equal to
ω cf = (1/ ((v π /2Rо) - v Rо))) ∫ 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 at the starting point B. It is obviously equal to v / Rо. 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 line R = kα passing through the origin has a limit. In this case, the limit does not exist, but there is a limit for a certain line. Let's find the coefficient k in the equation of a straight line passing through the origin.
At α = 0 ® R= 0, at α = π /2 ® R= Rо (Fig. 3), hence k = 2Rо/π , then formula (5) is transformed to a form that includes the first remarkable limit
ℓim (v π sinα /2Rо α) = (v π/2Rо) ℓim sinα/α = v π/2Rо. (4.6)
α ® 0 α ® 0
Now we substitute the value obtained from formulas (4.1), (4.3) and (4.4) into (4.2) and obtain
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°
Finally, the sum of the projections of the Coriolis force 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, assuming the acceleration to be constant
Fc = ∑F^ cp = 4ρ r 2 w 2 ((π /2.) +1) / t) ∫t 2 dt.
After calculations, we get
Fc ≈ 1.3ρ r 2 w 2 ((π /2.) +1)t 2 . (4.9)
Let r = 0.02m; w \u003d 5m / s 2; ρ \u003d 1000 kg / m 3; t = 1c, then the total average Coriolis force of inertia during the action of the positive acceleration of the circulating medium will be Fk ≈ 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, we divide the trajectory into two halves and insert a semicircle in the lower part, and a straight line in the upper part, as shown in Fig. 4.4, and direct the circulating medium along a pipe of radius r, curved in 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 circulating medium circuit
Fc = 2 ρπ r 2 v 2 . (4.10)
Thus, the centrifugal force does not depend on the radius R, but depends only on the integration angle (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 anything, it can be concluded 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 own line, which are parallel and do not lie on the same line.
The sum of 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 a broken line and projections acting along the X axis, arising in two convex curves under broken lines are compensated, since they are the same and directed in opposite directions. Thus. centrifugal force does not contribute to translational motion.
§5. Solid state rotational systems. Centrifugal forces of inertia.
1. The vector of the own angular velocity of the rods is perpendicular to the vector of the angular velocity 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 of rotational motion and vice versa. Consider a pair of opposite rods of length ℓ with point weights of the same 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. 5.1): half a turn of the rod in one revolution around a common axis. Let R³ℓ/2. For a complete description of the process, it suffices to consider rotation in the range of angles 0£ α £ π/2. We 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.
The angle α is related to frequency ω and time t by
α = ωt/2, (5.1.1)
since a half-turn of the rod occurs in one revolution around a common axis. It is clear that the centrifugal force inertia there will be more distant cargoes from the center than near ones. Projections of centrifugal forces inertia on the X axis will be
Fц1 = mω 2 (R - (ℓ/2) cos α) sin 2α (5.1.2)
Fц2 = mω 2 (R + (ℓ/2) cos α) sin 2α (5.1.3)
Fц3 = - mω 2 (R + (ℓ/2) sin α) sin 2α (5.1.4)
Fц4 = - mω 2 (R - (ℓ/2) sin α) sin 2α (5.1.5)
We write the difference centrifugal force inertia acting on remote loads. Difference centrifugal force inertia on the second load
Fц2-1 = mω 2 ℓ cosα sin2α. (5.1.6)
Difference centrifugal force inertia on the third load
Fц3-4 = - mω 2 ℓ sinα sin2α. (5.1.7)
Average value of differential centrifugal forces inertia for a half turn
Fav c2-1 = (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)
Received two opposite and equal in absolute value centrifugal forces inertia that are external. Therefore, they can be represented as two identical infinitely distant bodies (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 action on the system per half-turn along the X axis is equal to the sum of the forces of pulling Fav c2-1 and repulsion Fav c3-4 from external bodies
Fp = | Fcp c2-1 | + | Fav ts3-4 | = 0.8 mω 2 ℓ. (5.1.10)
To eliminate the torque of the system of two rods in the vertical plane (Fig. 5.2), it is necessary to apply 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 the 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 \u003d 4Fp \u003d 3.2mω 2 ℓ. (5.1.11)
Let m = 0.1kg; ω =2 πf, where f = 10r/s; ℓ = 0.5m, then Ft ≈ 632N.
2. The vector of the own angular velocity of the rods is perpendicular to the vector of the angular velocity of the center of mass of the rod and is parallel to the radius of the common axis of rotation of the rods.
Let us consider a pair of rods of length ℓ opposite perpendicular to each other with point weights of the same 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 turn of the rod in one revolution around a common axis.
Rice. 5.4.
For the 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 velocity 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.
We single out 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 is the distance from the center O to the center of mass of the loads, r1, r2 are the 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's find the total angular velocity of each load relative to the common axis of rotation, given that the linear velocities are oppositely directed 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 \u003d mω 2 2 r2
Or in detail
F 1 \u003d 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 \u003d 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)
Consider the case when ℓ=4R. In this case, atΨ=180° angular frequency of the first weight 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 the definition of centrifugal forces in the direction of the X axis at ℓ= 4R
F 1 \u003d 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 \u003d 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 ° centrifugal force projection 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 СР 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 СР 1-2 in the formula (5.2.12) is positive, that is, it 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 60° £ Ψ £180°
F СР 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 the interval 0° £ Ψ £180° will obviously be
F СР = (F СР 1-2 + 2F СР 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 \u003d 4 F СР \u003d 5.6mω 2 R. (5.2.15)
Let m = 0.1kg; ω =2 πf, where f = 10r/s; ℓ= 4R, where R = 0.1m, then F T ≈ 220N.
3. The vector of the own angular velocity of the rods is parallel and equally directed with the vector of the angular velocity of the center of mass of the rod rotating about a common axis.
Let us consider a pair of opposite, lying on the water plane, rods of length ℓ with point weights of the same 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 turn of the rod in one revolution around a common axis.
Rice. 5.7.
Similarly to the previous case, we choose only m1 and m2 for calculation, since the solution for m3 and m4 is similar. We will make an approximate estimate of the acting forces of inertia 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 through a half turn for both ω. The distance from the first weight to the center O at the beginning of 2R from the second weight is 0, and after a half turn from each RÖ 2.
Rice. 5.8.
And 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 body 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) 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 СР 1 = (ω + 0.5ω + ω)/2 = 1.25ω. (5.3.1)
Average angular velocity of the second load
ω СР 2 = (ω - 0.5ω + ω)/2 = 0.75ω. (5.3.2)
Average radius of the first load
R SR 1 = (2R + R Ö 2)/2 = R(2 + Ö 2)/2.(5.3.3)
Average radius of the second load
R СР 2 =(0 + R Ö 2)/2 = (RÖ 2)/2.(5.3.4)
The projection of the centrifugal force acting on the first weight in the direction of the X axis will be
F 1 = mω 2 СР 1 R СР 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 weight in the direction of the X axis will be
F 2 = mω 2 СР 2 R СР 2 sin(Ψ /2)sin2Ψ » 0.4mω 2 R sin(Ψ /2)sin2Ψ. (5.3.6)
° £ Ψ £36° will be
0.2p
F СР 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
0.4p
F СР 1 - 2 = (1/0.2 π) ∫ (F 1 - F 2) dΨ » 1.95mω 2 R. (5.3.8)
0.2p
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
0.5p
F СР- (1 + 2) \u003d - (1 / 0.1 π) ∫ (F 1 + F 2)dΨ "-3.72mω 2 R. (5.3.9)
0.4p
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 СР = (2F СР 1 + 2 + 2F СР 1 – 2 + F СР- (1 + 2))/5 » 0.62mω 2 R. (5.3.10)
Similarly, the sum of projections of centrifugal forces for the third and fourth loads is calculated.
To eliminate the torque, it is necessary to apply 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 thrust force will be
F T \u003d 4F СР \u003d 2.48mω 2 R. (5.3.11)
Let m = 0.1kg; ω =2 πf, where f = 10r/s; R = 0.25m, then F T ≈ 245N.
§6. Phase force of inertia.
To implement the phase inertia force, we use a two-crank articulated four-link as a translational one to convert the uniform rotation of the engine into uneven rotation of loads according to a certain mode with optimization of the nature of the movement of goods for the effective use of inertia forces, and by appropriate choice of the relative position of the loads, compensate for the reverse impulse
The articulated four-link will be two-crank if the center distance of the AG (Fig.6.1) will be less than the length of any movable link, and the sum of the center-to-center distance and the length of the largest of the movable links will be less than the sum of the lengths of the other two links.
Rice. 6.1.
Link VG (lever), on which a load of mass m is fixed, is a driven crank on a fixed shaft G, and link AB is the 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 driving crank is chosen so that when the load reaches the extreme point D, there is a right angle between the connecting rod and the driving crank, which ensures maximum efficiency. Then, with uniform rotation of the motor shaft A with the driving crank AB with an angular velocity w, the connecting rod BV transmits the 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 force of inertia acts in the direction of movement of the load. 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 inertia force acts against the direction of movement of the load, coinciding with the direction of the inertia force in the first semicircle. The integrated propulsion scheme 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 motor shaft, and the driven cranks (levers) independently rotate on a fixed shaft. The longitudinal components of the inertia forces in the direction from point E to point D of the upper and lower loads are added, providing translational motion. There is no reverse impulse, since the weights rotate in the same direction and, on average, are symmetrically opposite.
Let's estimate the acting phase force of inertia.
Let AB = BV = r, GV = R.
Suppose that in the extreme right position, the angle Ψ between the radius R and the middle line DE is 0° (Fig.6.4) and
r + r - AG = R, (6.1)
and also in the extreme left position at Ψ =180° (Fig.6.5) the angle
Ð ABV = 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+r 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 GV 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 HW = (180° /225° )w . (6.5)
The increment of the angular velocity ∆w during the time ∆t = 225° /w = 5π/4w will be
∆w = w GW - w = - 0.2w . (6.6)
Let the angular acceleration be equally slow, then
dω / dt \u003d ∆w / ∆t \u003d - 0.16w 2 / π. (6.7)
Let us use the formula of the phase force of inertia (2.8) in scalar form
F f \u003d -m [(dω / dt) R] \u003d 0.16mw 2 R / π. (6.8)
Rice. 6.5.
The projection of the phase force of inertia in the direction of ED will be
F FED \u003d 0.16mw 2 RsinΨ / π. (6.9)
The average value of the projection of the phase force of inertia for a half-cycle
F СР = 0.16mω 2 R/ π 2) ∫ sinΨdΨ = 0.32mω 2 R/ π 2 . (6.10)
For two loads (Fig. 6.3) the force is doubled. 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 \u003d 4F СР \u003d 1.28mω 2 R / π 2. (6.11)
Let m = 0.1kg; ω =2 πf, where f = 10r/s; R = 0.5m, then F T = 25.6N.
§7. Gyroscope. Coriolis and centrifugal force of inertia.
Consider the oscillatory movement of a load of mass m along a semicircle (Fig. 7.1) with radius R with a linear speed v. The centrifugal inertia force Fc acting on a load of mass m will be equal to m v 2 / R, directed along the radius from the center O. The projection of the centrifugal force on the X axis will be equal to
F c׀׀ \u003d (m v 2 / R) sin α. (7.1)
The load must move with acceleration w around the circumference so that the centrifugal force is effective for the translational movement 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 under the influence of a force that changes the direction of the gyroscope axis, it precesses under the influence of the Coriolis force, and this movement is inertialess. That is, with the instantaneous application of a force that changes the direction of the axis of rotation, the gyroscope instantly begins to precess and just as instantly stops when this force disappears. Instead of a load, we use a gyroscope rotating at an angular velocity ω. Now we apply force F perpendicular to the axis of rotation of the gyroscope (Fig. 7.2) and act on the axis so that the holder with the gyroscope performs an inertial oscillatory motion (precesses) in a certain sector (in the optimal case with a final value of α = 180 °). The instantaneous stop of the precession of the holder with the gyroscope and its resumption in the opposite direction occurs when the direction of the force F changes to the opposite. Thus, there is an oscillatory inertialess movement of the holder with a gyroscope, which eliminates the reverse impulse that prevents translational movement along the X axis.
Rice. 7.2.
Angular speed of precession
dα /dt = M / I Z ω, (7.3)
where: M - moment of force; I Z is the moment of inertia of the gyroscope; ω is the angular velocity of the gyroscope.
Moment of force (assuming ℓ is perpendicular to F)
M = ℓ F, (7.4)
where: ℓ is the distance from the point of application of the force F to the center of inertia of the gyroscope; F is the 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 , ω are considered constant, and the force F, depending on the time t, let it change according to a piecewise linear law (Fig. 7.3).
Rice. 7.3.
It is known that the linear velocity is related to the angular velocity by the following relation
v = R (dα /dt). (7.6)
Differentiating formula (7.6) with respect to time, we obtain the acceleration
w = R (d 2 α /dt 2). (7.7)
We substitute formula (7.5) into formula (7.7) and obtain
w = (R ℓ/I Zω ) (dF/dt). (7.8)
Thus, the acceleration depends on the rate of change of the force F, which makes the centrifugal force acting 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 Fц ┴, we use the second same gyroscope, which oscillates 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 one. It is obvious that the perpendicular components Fц ┴ will be compensated, and the parallel ones Fц׀׀ will add up.
Rice. 7.4.
If the oscillation sector of the gyroscopes is not more than a semicircle, then there will be no opposite centrifugal force, which reduces the centrifugal force in the direction of the X axis.
To eliminate the torque of the device, which occurs due to the forced rotation of the axis of the gyroscopes, it is necessary to install another pair of the same gyroscopes, the axes of which rotate in the opposite direction. Sectors of oscillatory motion of holders with gyroscopes in a pair, the gyroscope axes of which rotate in one direction, must be symmetrically directed in one direction with sectors of holders with gyroscopes, the gyroscope axes of which rotate in the opposite direction (Fig. 7.5).
Rice. 7.5.
Let us calculate the average value of the centrifugal force projection Fц׀׀ for one gyroscope (Fig. 7.2) on the holder, oscillating in the sector of the semicircle 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 Fp 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 = 1kg. 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 a small interval of absence of the centrifugal force in the center. Speed up 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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