Lectures on theoretical mechanics
Point dynamics
Lecture 1
Basic concepts of dynamics
In chapter Dynamics the movement of bodies under the action of forces applied to them is studied. Therefore, in addition to the concepts that were introduced in the section Kinematics, here it is necessary to use new concepts that reflect the specifics of the effect of forces on various bodies and the reaction of bodies to these influences. Let's consider the main of these concepts.
a) strength
Force is a quantitative result of the impact on a given body from other bodies. Force is a vector quantity (Fig. 1).
Point A of the beginning of the force vector F called point of application of force... The straight line MN on which the force vector is located is called line of action of force. The length of a force vector, measured on a certain scale, is called numerical value or modulus of the force vector... The force modulus is denoted as or. The action of the force on the body is manifested either in its deformation, if the body is motionless, or in imparting acceleration to it when the body moves. On these manifestations of force, the device of various devices (force meters or dynamometers) for measuring forces is based.
b) system of forces
The set of forces under consideration forms system of forces. Any system consisting of n forces can be written in the following form: 
c) free body
A body that can move in space in any direction without experiencing direct (mechanical) interaction with other bodies is called free or isolated... The effect of one or another system of forces on a body can be clarified only if this body is free.
d) resultant force
If any force exerts the same effect on a free body as a certain system of forces, then this force is called the resultant of this system of forces... This is written as follows:
,
which means equivalence action on one and the same free body of the resultant and some system of n forces.
Let us now proceed to consider more complex concepts related to the quantitative determination of the rotational effects of forces.
e) moment of force about a point (center)
If the body under the action of force can rotate around some fixed point O (Fig. 2), then to quantify this rotational effect, a physical quantity is introduced, which is called moment of force about a point (center).
The plane passing through a given fixed point and the line of action of the force is called plane of action of force... In Fig. 2, this is the plane ОАВ.
The moment of force relative to a point (center) is a vector quantity equal to the vector product of the radius vector of the point of application of the force by the force vector:
( 1)
According to the rule of vector multiplication of two vectors, their vector product is a vector perpendicular to the plane of location of the vectors of the factors (in this case, the plane of the triangle OAB), directed in the direction from which the shortest rotation of the first vector of the factor to the second vector is the factor visible against the clock hand (Fig. 2). With this order of the vectors of the factors of the vector product (1), the rotation of the body under the action of the force will be visible against the clock hand (Fig. 2) Since the vector is perpendicular to the plane of action of the force, its location in space determines the position of the plane of action of the force. relative to the center is equal to the doubled area ОАВ and can be determined by the formula:
,
(2)
where magnitudeh, equal to the shortest distance from a given point O to the line of action of the force, is called the shoulder of the force.
If the position of the plane of action of the force in space is not essential for the characteristic of the rotational action of the force, then in this case, to characterize the rotational action of the force, instead of the vector of the moment of force is used algebraic moment of force:
(3)
The algebraic moment of force relative to a given center is equal to the product of the modulus of force by its shoulder, taken with a plus or minus sign. In this case, the positive moment corresponds to the rotation of the body under the action of the given force against the clock hand, and the negative moment corresponds to the rotation of the body along the clock hand. From formulas (1), (2) and (3) it follows that the moment of force relative to the point is zero only if the shoulder of this forcehequal to zero... Such a force cannot rotate the body around a given point.
f) Moment of force about the axis
If the body under the action of force can rotate around some fixed axis (for example, the rotation of a door or window frame in hinges when they are opened or closed), then to quantify this rotational effect, a physical quantity is introduced, which is called moment of force about a given axis.
z
b F xy
Figure 3 shows a diagram in accordance with which the moment of force relative to the z axis is determined:
The angle is formed by two perpendicular directions z and to the planes of triangles O ab and OAV, respectively. Since O ab is the projection of ОАВ onto the xy plane, then by the stereometry theorem on the projection of a plane figure onto this plane we have:
where the plus sign corresponds to the positive value of cos, i.e. acute angles , and the minus sign corresponds to the negative value of cos, i.e. to obtuse angles , which is due to the direction of the vector. In turn, SO ab=1/2abh, where h ab ... Segment size ab is equal to the projection of the force on the xy plane, i.e. . ab = F xy .
Based on the above, as well as equalities (4) and (5), we define the moment of force relative to the z axis as follows:
Equality (6) allows us to formulate the following definition of the moment of force relative to any axis: The moment of force relative to a given axis is equal to the projection on this axis of the vector of the moment of this force relative to any point of this axis and is defined as the product of the projection of the force on the plane perpendicular to this axis, taken with a plus or minus sign on the shoulder of this projection relative to the point of intersection of the axis with the projection plane. In this case, the sign of the moment is considered positive if, looking from the positive direction of the axis, the rotation of the body around this axis is visible against the clock hand. Otherwise, the moment of force about the axis is taken negative. Since this definition of the moment of force about the axis is rather difficult to memorize, it is recommended to remember the formula (6) and Fig. 3, which explains this formula.
From formula (6) it follows that the moment of force about the axis is zero if it is parallel to the axis (in this case, its projection onto a plane perpendicular to the axis is zero), or the line of action of the force intersects the axis (then the shoulder of the projection h=0). This fully corresponds to the physical meaning of the moment of force about the axis as a quantitative characteristic of the rotational effect of the force on a body having an axis of rotation.
g) body weight
It has long been noticed that under the action of force, a body gains speed gradually and continues to move if the force is removed. This property of bodies, to resist a change in their movement, was called inertia or inertia of bodies. The quantitative measure of the inertness of a body is its mass. Besides, body mass is a quantitative measure of the effect of gravitational forces on a given body the greater the mass of the body, the greater the gravitational force acting on the body. As shown below, NS These two definitions of body weight are related.
The rest of the concepts and definitions of dynamics will be discussed later in the sections where they first appear.
2. Bonds and bond reactions
Earlier in section 1, point (c), the concept of a free body was given, as a body that can move in space in any direction without being in direct contact with other bodies. Most of the real bodies that surround us are in direct contact with other bodies and cannot move in one direction or another. So, for example, bodies on the table surface can move in any direction, except for the direction perpendicular to the table surface downward. Hinged doors can rotate, but cannot translate, etc. Bodies that cannot move in space in one direction or another are called not free.
Anything that limits the movement of a given body in space is called constraints. It can be any other bodies that prevent the movement of this body in some directions ( physical connections); in a broader sense, it can be some conditions imposed on the movement of the body, limiting this movement. So, you can put a condition that the movement of a material point occurs along a given curve. In this case, the connection is specified mathematically in the form of an equation ( constraint equation). The question of the types of links will be discussed in more detail below.
Most of the connections imposed on bodies are practically physical connections. Therefore, the question arises about the interaction of this body and the connection imposed on this body. This question is answered by the axiom about the interaction of bodies: Two bodies act on each other with forces equal in magnitude, opposite in direction and located on the same straight line. These forces are called interaction forces. Interaction forces are applied to different interacting bodies. So, for example, when a given body and a connection interact, one of the interaction forces is applied from the side of the body to the connection, and the other interaction force is applied from the side of the connection to this body. This last power is called by the strength of the bond reaction or simply, communication reaction.
When solving practical problems of dynamics, it is necessary to be able to find the direction of reactions of various types of connections. The general rule of determining the direction of the bond reaction can sometimes help in this: The bond reaction is always directed opposite to the direction in which this bond prevents the movement of the given body. If this direction can be indicated definitely, then the reaction of the connection will be determined by the direction. Otherwise, the direction of the bond reaction is uncertain and can be found only from the corresponding equations of motion or equilibrium of the body. In more detail, the question of the types of connections and the direction of their reactions should be studied in the textbook: S.M. Targ A short course in theoretical mechanics "High school", M., 1986. Chapter 1, §3.
In section 1, point (c), it was said that the effect of any system of forces can be fully determined only if this system of forces is applied to a free body. Since most bodies, in reality, are not free, then in order to study the movement of these bodies, the question arises of how to make these bodies free. This question is answered axiom of lecture connections on philosophy at home. Lectures were ... social psychology and ethnopsychology. 3. Theoretical Outcomes In social Darwinism there were ...
Theoretical Mechanics
Study Guide >> PhysicsAbstract lectures on subject THEORETICAL MECHANICS For students of the specialty: 260501.65 ... - full-time Abstract lectures compiled on the basis of: L.V. Butorin, E.B. Busygin. Theoretical Mechanics... Training manual ...
In any academic course, the study of physics begins with mechanics. Not with theoretical, not with applied and not computational, but with good old classical mechanics. This mechanics is also called Newtonian mechanics. According to legend, the scientist was walking in the garden, saw an apple fall, and it was this phenomenon that pushed him to the discovery of the law of universal gravitation. Of course, the law has always existed, and Newton only gave it a form that people understand, but his merit is priceless. In this article, we will not describe the laws of Newtonian mechanics in as much detail as possible, but we will outline the basics, basic knowledge, definitions and formulas that can always play into your hands.
Mechanics is a branch of physics, a science that studies the movement of material bodies and the interactions between them.
The word itself is of Greek origin and is translated as "the art of building machines." But before the construction of machines, we are still like the Moon, so we will follow in the footsteps of our ancestors, and we will study the movement of stones thrown at an angle to the horizon, and apples falling on heads from a height of h.
Why does the study of physics begin with mechanics? Because it is completely natural, not to start it from thermodynamic equilibrium ?!
Mechanics is one of the oldest sciences, and historically the study of physics began precisely from the foundations of mechanics. Placed within the framework of time and space, people, in fact, could not start from something else, with all their desire. Moving bodies are the first thing we turn our attention to.
What is movement?
Mechanical movement is a change in the position of bodies in space relative to each other over time.
It is after this definition that we quite naturally come to the concept of a frame of reference. Changing the position of bodies in space relative to each other. Key words here: relative to each other ... After all, a passenger in a car moves relative to a person standing on the side of the road at a certain speed, and rests relative to his neighbor on the seat next to him, and moves at some other speed relative to a passenger in a car that overtakes them.
That is why, in order to normally measure the parameters of moving objects and not get confused, we need frame of reference - rigidly interconnected reference body, coordinate system and clock. For example, the earth moves around the sun in a heliocentric frame of reference. In everyday life, we carry out almost all of our measurements in a geocentric frame of reference associated with the Earth. Earth is a reference body, relative to which cars, airplanes, people, animals move.
Mechanics, as a science, has its own task. The task of mechanics is to know the position of a body in space at any time. In other words, mechanics constructs a mathematical description of motion and finds connections between the physical quantities that characterize it.
In order to move further, we need the concept “ material point ”. They say physics is an exact science, but physicists know how many approximations and assumptions have to be made in order to agree on this very accuracy. Nobody has ever seen a material point or smelled ideal gas, but they are! It's just much easier to live with them.
Material point is a body, the size and shape of which can be neglected in the context of this problem.
Sections of classical mechanics
Mechanics consists of several sections
- Kinematics
- Dynamics
- Statics
Kinematics from a physical point of view, it studies exactly how the body moves. In other words, this section deals with the quantitative characteristics of movement. Find speed, path - typical kinematic problems
Dynamics solves the question of why it moves that way. That is, it considers the forces acting on the body.
Statics studies the balance of bodies under the action of forces, that is, answers the question: why does it not fall at all?
The limits of applicability of classical mechanics
Classical mechanics no longer claims to be a science that explains everything (at the beginning of the last century, everything was completely different), and has a clear framework of applicability. In general, the laws of classical mechanics are valid for the world we are accustomed to in terms of size (macrocosm). They stop working in the case of the particle world, when quantum mechanics replaces the classical one. Also, classical mechanics is inapplicable to cases when the movement of bodies occurs at a speed close to the speed of light. In such cases, relativistic effects become pronounced. Roughly speaking, within the framework of quantum and relativistic mechanics - classical mechanics, this is a special case when the dimensions of the body are large, and the speed is small.
Generally speaking, quantum and relativistic effects never go anywhere; they also take place during the ordinary motion of macroscopic bodies with a speed much less than the speed of light. Another thing is that the effect of these effects is so small that it does not go beyond the most accurate measurements. Thus, classical mechanics will never lose their fundamental importance.
We will continue to study the physical foundations of mechanics in future articles. For a better understanding of the mechanics, you can always refer to to our authors who individually shed light on the dark spot of the most difficult task.
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Course of lectures on theoretical mechanics Dynamics (I part) Bondarenko A.N. Moscow - 2007 The electronic training course was written on the basis of the lectures given by the author for students who studied in the specialties of the SZD, PGS and SDM at NIIZhT and MIIT (1974-2006). The educational material corresponds to the calendar plans in the volume of three semesters. To fully implement animation effects during a presentation, you must use a Power Point viewer no lower than the one built into Microsoft Office of the Windows-XP Professional operating system. Comments and suggestions can be sent by e-mail: [email protected]... Moscow State University of Railway Engineering (MIIT) Department of Theoretical Mechanics Scientific and Technical Center for Transport Technologies
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Contents Lecture 1. Introduction to dynamics. Laws and axioms of the dynamics of a material point. Basic equation of dynamics. Differential and natural equations of motion. Two main tasks of dynamics. Examples of solving the direct problem of dynamics Lecture 2. Solution of the inverse problem of dynamics. General instructions for solving the inverse problem of dynamics. Examples of solving the inverse problem of dynamics. The movement of a body thrown at an angle to the horizon, disregarding air resistance. Lecture 3. Rectilinear vibrations of a material point. Condition for the occurrence of vibrations. Classification of vibrations. Free vibrations without taking into account the forces of resistance. Damped Oscillations. Decrement of fluctuations. Lecture 4. Forced oscillations of a material point. Resonance. The influence of resistance to movement during forced vibrations. Lecture 5. Relative motion of a material point. Forces of inertia. Particular cases of movement for various types of portable movement. The influence of the Earth's rotation on the balance and motion of bodies. Lecture 6. Dynamics of a mechanical system. Mechanical system. External and internal forces. The center of mass of the system. The theorem on the motion of the center of mass. Conservation laws. An example of solving a problem using the theorem on the motion of the center of mass. Lecture 7. Impulse of power. The amount of movement. The theorem on the change in the amount of motion. Conservation laws. Euler's theorem. An example of solving the problem of using the theorem about changing the momentum. Moment of momentum. Theorem about the change in the angular momentum .. Lecture 8. Conservation laws. Elements of the theory of moments of inertia. Kinetic moment of a rigid body. Differential equation of rotation of a rigid body. An example of solving the problem on the use of the theorem on the change in the angular momentum of the system. Elementary theory of the gyroscope. Recommended reading 1. Yablonskiy A.A. Course of theoretical mechanics. Part 2. M .: Higher school. 1977 368 s. 2. Meshchersky I.V. Collection of problems in theoretical mechanics. M .: Science. 1986 416 s. 3. Collection of assignments for term papers / Ed. A.A. Yablonsky. M.: Higher school. 1985 366 p. 4. Bondarenko A. N. “Theoretical mechanics in examples and problems. Dynamics ”(electronic manual www.miit.ru/institut/ipss/faculties/trm/main.htm), 2004
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Lecture 1 Dynamics is a branch of theoretical mechanics that studies mechanical motion from the most general point of view. The movement is considered in connection with the forces acting on the object. The section consists of three sections: Dynamics of a material point Dynamics Dynamics of a mechanical system Analytical mechanics ■ Dynamics of a point - studies the movement of a material point taking into account the forces causing this movement. The main object is a material point - a material body with a mass, the dimensions of which can be neglected. Basic assumptions: - there is an absolute space (it has purely geometric properties that do not depend on matter and its motion. - there is an absolute time (does not depend on matter and its motion). Hence follows: - there is an absolutely motionless frame of reference. - time does not depend on the motion of the frame of reference. - the masses of moving points do not depend on the motion of the frame of reference. These assumptions are used in classical mechanics, created by Galileo and Newton. It still has a fairly wide field of application, since the mechanical systems considered in applied sciences do not have such large masses and speeds of motion, for which it is necessary to take into account their influence on the geometry of space, time, motion, as is done in relativistic mechanics (theory of relativity). their dynamic interaction Actions under the influence of various forces. ■ The law of inertia (Galileo-Newton's law) - An isolated material point, the body retains its state of rest or uniform rectilinear motion until the applied forces force it to change this state. This implies the equivalence of the state of rest and motion by inertia (Galileo's law of relativity). The frame of reference in relation to which the law of inertia is fulfilled is called inertial. The property of a material point to strive to keep the speed of its movement (its kinematic state) unchanged is called inertia. ■ The law of proportionality of force and acceleration (Basic equation of dynamics - Newton's II law) - The acceleration imparted to a material point by force is directly proportional to the force and inversely proportional to the mass of this point: or Here m is the mass of the point (measure of inertia), measured in kg, numerically equal weight divided by the acceleration due to gravity: F is the acting force, measured in N (1 N imparts an acceleration of 1 m / s2 to a point with a mass of 1 kg, 1 N = 1/9. 81 kg-s). ■ Dynamics of a mechanical system - studies the motion of a set of material points and solids, united by general laws of interaction, taking into account the forces that cause this motion. ■ Analytical Mechanics - studies the motion of non-free mechanical systems using general analytical methods. 1
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Lecture 1 (continued - 1.2) Differential equations of motion of a material point: - differential equation of motion of a point in vector form. - differential equations of motion of a point in coordinate form. This result can be obtained by formal projection of the vector differential equation (1). After grouping, the vector relation breaks down into three scalar equations: In coordinate form: We use the relationship of the radius vector with coordinates and the force vector with projections: or: Substitute the acceleration of a point in the vector setting of motion into the basic equation of dynamics: Natural equations of motion of a material point are obtained by projecting a vector differential equation of motion on natural (movable) coordinate axes: or: - natural equations of motion of a point. ■ The basic equation of dynamics: - corresponds to the vector method of specifying the motion of a point. ■ The law of independence of the action of forces - The acceleration of a material point under the action of several forces is equal to the geometric sum of the accelerations of a point from the action of each of the forces separately: or the Law is valid for any kinematic state of bodies. The forces of interaction, being applied to different points (bodies), are not balanced. ■ The law of equality of action and reaction (III Newton's law) - Every action corresponds to an equal in magnitude and oppositely directed reaction: 2
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Two main problems of dynamics: 1. Direct problem: Motion is given (equations of motion, trajectory). It is required to determine the forces under the action of which a given movement occurs. 2. Inverse problem: The forces under the action of which the movement occurs. It is required to find the parameters of motion (equations of motion, trajectory of motion). Both problems are solved using the basic equation of dynamics and its projection onto the coordinate axes. If the motion of a non-free point is considered, then, as in statics, the principle of freedom from bonds is used. As a result of the reaction, the bonds are included in the composition of the forces acting on the material point. The solution to the first problem is associated with differentiation operations. The solution of the inverse problem requires the integration of the corresponding differential equations, and this is much more difficult than differentiation. The inverse problem is more complicated than the direct problem. Let us consider the solution of the direct problem of dynamics by examples: Example 1. An elevator car with weight G is lifted by a cable with acceleration a. Determine the cable tension. 1. We select an object (the elevator car moves progressively and it can be considered as a material point). 2. We discard the connection (cable) and replace with reaction R. 3. Make up the basic equation of dynamics: Determine the reaction of the cable: Determine the tension of the cable: With uniform movement of the cab ay = 0 and the tension of the cable is equal to the weight: T = G. When the cable breaks, T = 0 and the acceleration of the cabin is equal to the acceleration of gravity: ay = -g. 3 4. Let's project the basic equation of dynamics on the y-axis: y Example 2. A point of mass m moves along a horizontal surface (Oxy plane) according to the equations: x = a coskt, y = b coskt. Determine the force acting on the point. 1. Select an object (material point). 2. We discard the connection (plane) and replace it with reaction N. 3. Add the unknown force F. to the system of forces 4. We compose the basic equation of dynamics: 5. We project the basic equation of dynamics on the x, y axes: We determine the projections of the force: Modulus of force: Direction cosines : Thus, the magnitude of the force is proportional to the distance of the point from the center of coordinates and is directed towards the center along the line connecting the point to the center. The trajectory of a point is an ellipse centered at the origin: O r Lecture 1 (continued - 1.3)
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Lecture 1 (continued 1.4) Example 3: A load of weight G is suspended on a cable of length l and moves along a circular path in a horizontal plane at a certain speed. The angle of deviation of the cable from the vertical is equal. Determine the rope tension and load speed. 1. Select the object (cargo). 2. We discard the connection (cable) and replace with reaction R. 3. Make up the basic equation of dynamics: From the third equation we determine the reaction of the cable: We determine the tension of the cable: Substitute the value of the cable reaction, normal acceleration into the second equation and determine the speed of the load: 4. Project the basic equation dynamics on the axle, n, b: Example 4: A car with weight G moves along a convex bridge (radius of curvature is R) at a speed V. Determine the pressure of the car on the bridge. 1. We select an object (a car, we neglect the dimensions and consider it as a point). 2. We discard the bond (rough surface) and replace with reactions N and friction force Ffr. 3. We compose the basic equation of dynamics: 4. We project the basic equation of dynamics on the axis n: From here we determine the normal reaction: We determine the pressure of the car on the bridge: From here we can determine the speed corresponding to zero pressure on the bridge (Q = 0): 4
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Lecture 2 After substitution of the found values of the constants, we obtain: Thus, under the action of the same system of forces, a material point can perform a whole class of motions determined by the initial conditions. The origin coordinates take into account the origin of the point. The initial velocity, given by the projections, takes into account the effect on its movement along the considered section of the trajectory of the forces acting on the point before arriving at this section, i.e. initial kinematic state. Solution of the inverse problem of dynamics - In the general case, the motion of a point of force acting on a point are variables that depend on time, coordinates and speed. The motion of a point is described by a system of three second-order differential equations: After integrating each of them, there will be six constants C1, C2,…., C6: The values of the constants C1, C2,…., C6 are found from six initial conditions at t = 0: Solution example 1 inverse problem: A free material point of mass m moves under the action of a force F, constant in magnitude and magnitude. ... At the initial moment, the speed of the point was v0 and coincided in direction with the force. Determine the equation of motion of a point. 1. Make up the basic equation of dynamics: 3. Lower the order of the derivative: 2. Choose a Cartesian frame of reference, directing the x-axis along the direction of the force and project the basic equation of dynamics on this axis: or xyz 4. Separate the variables: 5. Calculate the integrals of both sides of the equation : 6. We represent the projection of the velocity as the derivative of the coordinate with respect to time: 8. We calculate the integrals of both sides of the equation: 7. Separate the variables: 9. To determine the values of the constants C1 and C2, we use the initial conditions t = 0, vx = v0, x = x0: As a result, we get the equation of uniform motion (along the x-axis): 5
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General instructions for solving the direct and inverse problem. Solution procedure: 1. Compilation of the differential equation of motion: 1.1. Select a coordinate system - rectangular (fixed) with an unknown trajectory of motion, natural (movable) with a known trajectory, for example, a circle or a straight line. In the latter case, one straight-line coordinate can be used. Align the origin with the initial position of the point (at t = 0) or with the equilibrium position of the point, if it exists, for example, when the point vibrates. 6 1.2. Draw a point in a position corresponding to an arbitrary moment in time (for t> 0) so that the coordinates are positive (s> 0, x> 0). In this case, we also assume that the projection of the velocity in this position is also positive. In the case of oscillations, the projection of the velocity changes sign, for example, upon returning to the equilibrium position. Here it should be assumed that at the considered moment of time, the point moves away from the equilibrium position. This recommendation is important in the future when working with speed-dependent resistance forces. 1.3. Free the material point from connections, replace their action with reactions, add active forces. 1.4. Write down the basic law of dynamics in vector form, project onto selected axes, express the given or reactive forces in terms of time variables, coordinates or velocities, if they depend on them. 2. Solution of differential equations: 2.1. Lower the derivative if the equation is not reduced to the canonical (standard) form. for example: or 2.2. Split variables, for example: or 2.4. Calculate indefinite integrals on the left and right sides of the equation, for example: 2.3. If there are three variables in the equation, then make a change of variables, for example: and then divide the variables. Comment. Instead of computing indefinite integrals, you can compute definite integrals with a variable upper limit. The lower limits represent the initial values of the variables (initial conditions). Then a separate determination of the constant is not required, which is automatically included in the solution, for example: Using the initial conditions, for example, t = 0, vx = vx0, determine the constant of integration: 2.5. Express the speed in terms of the derivative of the coordinates in time, for example, and repeat paragraphs 2.2 - 2.4. Note. If the equation is reduced to the canonical form, which has a standard solution, then this ready-made solution is used. Integration constants are still found from the initial conditions. See, for example, hesitation (Lecture 4, p. eight). Lecture 2 (continued 2.2)
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Lecture 2 (continued 2.3) Example 2 of solving the inverse problem: Force depends on time. A load of weight P begins to move on a smooth horizontal surface under the action of a force F, the value of which is proportional to time (F = kt). Determine the distance traveled by the load in time t. 3. Make up the basic equation of dynamics: 5. Lower the order of the derivative: 4. Project the basic equation of dynamics on the x-axis: or 7 6. Separate the variables: 7. Calculate the integrals of both sides of the equation: 9. Let us represent the projection of the velocity as the time derivative of the coordinate: 10. Calculate the integrals of both sides of the equation: 9. Separate the variables: 8. Determine the value of the constant C1 from the initial condition t = 0, vx = v0 = 0: As a result, we obtain the equation of motion (along the x-axis), which gives the value of the distance traveled for time t: 1. We choose a frame of reference (Cartesian coordinates) so that the body has a positive coordinate: 2. We take the object of motion as a material point (the body moves translationally), release it from the connection (reference plane) and replace it with a reaction (normal reaction of a smooth surface) : 11. Determine the value of the constant C2 from the initial condition t = 0, x = x0 = 0: Example 3 of solving the inverse problem: The force depends on the coordinate. A material point of mass m is thrown upward from the surface of the Earth with a speed of v0. The Earth's gravity is inversely proportional to the square of the distance from a point to the center of gravity (the center of the Earth). Determine the dependence of the speed on the distance y to the center of the Earth. 1. We choose a frame of reference (Cartesian coordinates) so that the body has a positive coordinate: 2. We compose the basic equation of dynamics: 3. We project the basic equation of dynamics on the y-axis: or The coefficient of proportionality can be found using the weight of a point on the surface of the Earth: R Hence the differential the equation looks like: or 4. Lower the order of the derivative: 5. Make the change of variable: 6. Separate the variables: 7. Calculate the integrals of both sides of the equation: 8. Substitute the limits: As a result, we get the expression for the speed as a function of the y-coordinate: Maximum height flight speed can be found by equating the speed to zero: Maximum flight altitude when the denominator vanishes: Hence, when setting the Earth's radius and gravitational acceleration, the II cosmic speed is obtained:
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Lecture 2 (continuation 2.4) Example 2 of solving the inverse problem: Force depends on speed. A ship of mass m had a speed of v0. The resistance of the water to the movement of the vessel is proportional to the speed. Determine the time during which the boat's speed will drop by half after turning off the engine, as well as the distance traveled by the boat to a complete stop. 8 1. We choose a frame of reference (Cartesian coordinates) so that the body has a positive coordinate: 2. We take the object of motion as a material point (the ship is moving forward), release it from bonds (water) and replace it with a reaction (buoyant force - the force of Archimedes), and also by the force of resistance to movement. 3. Add active force (gravity). 4. Make up the basic equation of dynamics: 5. Project the basic equation of dynamics on the x-axis: or 6. Lower the order of the derivative: 7. Separate the variables: 8. Calculate the integrals of both sides of the equation: 9. Substitute the limits: An expression is obtained that relates the speed and time t, from where you can determine the time of movement: The time of movement, during which the speed will drop by half: It is interesting to note that when the speed approaches zero, the time of movement tends to infinity, i.e. final speed cannot be zero. Isn't it "perpetual motion"? However, the distance traveled to the stop is the final value. To determine the distance traveled, we turn to the expression obtained after lowering the order of the derivative, and make the change of the variable: After integration and substitution of the limits, we obtain: The distance traveled to the stop: ■ The movement of a point thrown at an angle to the horizon in a uniform gravity field without taking into account air resistance Eliminating time from the equations of motion, we obtain the trajectory equation: The flight time is determined by equating the y coordinate to zero: The flight range is determined by substituting the flight time:
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Lecture 3 Rectilinear oscillations of a material point - Oscillatory motion of a material point occurs under the condition: there is a restoring force that tends to return the point to the equilibrium position for any deviation from this position. 9 There is a restoring force, an equilibrium position is stable There is no restoring force, an equilibrium position is unstable There is no restoring force, an equilibrium position is indifferent. It is always directed to the equilibrium position, the value is directly proportional to the linear elongation (shortening) of the spring, equal to the deviation of the body from the equilibrium position: c is the coefficient of spring stiffness, numerically equal to the force under which the spring changes its length by one, measured in N / m in the system SI. x y O Types of vibrations of a material point: 1. Free vibrations (without taking into account the resistance of the medium). 2. Free vibrations taking into account the resistance of the medium (damped vibrations). 3. Forced vibrations. 4. Forced vibrations taking into account the resistance of the medium. ■ Free vibrations - occur under the influence of only the restoring force. Let us write down the basic law of dynamics: Choose a coordinate system centered at the equilibrium position (point O) and project the equation onto the x-axis: Let us reduce the resulting equation to the standard (canonical) form: This equation is a homogeneous linear differential equation of the second order, the form of solution of which is determined by the roots of the characteristic the equation obtained using the universal substitution: The roots of the characteristic equation are imaginary and equal: The general solution of the differential equation has the form: Point velocity: Initial conditions: Define the constants: So, the equation of free oscillations has the form: The equation can be represented by a single-term expression: - the initial phase. The new constants a and - are related to the constants C1 and C2 by the relations: Let us determine a and: The cause of the occurrence of free oscillations is the initial displacement x0 and / or the initial velocity v0.
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10 Lecture 3 (continued 3.2) Damped oscillations of a material point - Oscillatory motion of a material point occurs in the presence of a restoring force and a force of resistance to motion. The dependence of the force of resistance to motion on displacement or speed is determined by the physical nature of the medium or connection that prevents motion. The simplest dependence is a linear dependence on velocity (viscous resistance): - viscosity coefficient xy O Basic equation of dynamics: Projection of the equation of dynamics on the axis: Let us bring the equation to a standard form: where The characteristic equation has roots: The general solution of this differential equation has a different form depending on on the values of the roots: 1.n< k – случай малого вязкого сопротивления: - корни комплексные, различные. или x = ae-nt x = -ae-nt Частота затухающих колебаний: Период: T* Декремент колебаний: ai ai+1 Логарифмический декремент колебаний: Затухание колебаний происходит очень быстро. Основное влияние силы вязкого сопротивления – уменьшение амплитуды колебаний с течением времени. 2. n >k - case of high viscous resistance: - real roots, different. or - these functions are aperiodic: 3. n = k: - real multiple roots. these functions are also aperiodic:
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Lecture 3 (continuation 3.3) Classification of solutions of free oscillations. Springs connection methods. Equivalent stiffness. y y 11 Diff. equation Character. equation Roots character. equations Solution of a differential equation Graph nk n = k
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Lecture 4 Forced vibrations of a material point - Along with the restoring force, a periodically changing force, called the disturbing force, acts. The disturbing force can be of a different nature. For example, in a particular case, the inertial effect of the unbalanced mass m1 of a rotating rotor causes harmonically changing projections of the force: Basic equation of dynamics: Projection of the equation of dynamics onto an axis: Let us bring the equation to the standard form: 12 The solution of this inhomogeneous differential equation consists of two parts x = x1 + x2: x1 is the general solution of the corresponding homogeneous equation and x2 is the particular solution of the inhomogeneous equation: We select the particular solution in the form of the right-hand side: The obtained equality must be satisfied for any t. Then: or Thus, with the simultaneous action of the restoring and disturbing forces, the material point performs a complex oscillatory motion, which is the result of the addition (superposition) of free (x1) and forced (x2) oscillations. If p< k (вынужденные колебания малой частоты), то фаза колебаний совпадает с фазой возмущающей силы: В итоге полное решение: или Общее решение: Постоянные С1 и С2, или a и определяются из начальных условий с использованием полного решения (!): Таким образом, частное решение: Если p >k (forced oscillations of high frequency), then the phase of oscillations is opposite to the phase of the disturbing force:
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Lecture 4 (continued 4.2) 13 The dynamic coefficient is the ratio of the amplitude of forced vibrations to the static deflection of a point under the action of a constant force H = const: Amplitude of forced vibrations: The static deflection can be found from the equilibrium equation: Here: Hence: Thus, at p< k (малая частота вынужденных колебаний) коэффициент динамичности: При p >k (high frequency of forced vibrations) dynamic factor: Resonance - occurs when the frequency of forced vibrations coincides with the frequency of natural vibrations (p = k). This most often occurs when starting and stopping rotation of poorly balanced rotors attached to elastic suspensions. The differential equation of oscillations with equal frequencies: The particular solution in the form of the right-hand side cannot be taken, since you get a linearly dependent solution (see general solution). General solution: Substitute in the differential equation: Take a particular solution in the form and calculate the derivatives: Thus, the solution is obtained: or Forced oscillations at resonance have an amplitude that increases indefinitely in proportion to time. The influence of resistance to movement during forced vibrations. The differential equation in the presence of viscous resistance has the form: The general solution is selected from the table (Lecture 3, page 11), depending on the ratio of n and k (see). We take the particular solution in the form and calculate the derivatives: Substitute in the differential equation: Equating the coefficients for the same trigonometric functions, we obtain a system of equations: Raising both equations to the power and adding them to the power of both equations, we obtain the amplitude of the forced oscillations: By dividing the second equation by the first, we obtain the phase shift of the forced oscillations: Thus , the equation of motion for forced vibrations, taking into account the resistance to motion, for example, for n< k (малое сопротивление): Вынужденные колебания при сопротивлении движению не затухают. Частота и период вынужденных колебаний равны частоте и периоду изменения возмущающей силы. Коэффициент динамичности при резонансе имеет конечную величину и зависит от соотношения n и к.
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Lecture 5 Relative motion of a material point - Suppose that the moving (non-inertial) coordinate system Oxyz moves according to a certain law relative to the stationary (inertial) coordinate system O1x1y1z1. The motion of the material point M (x, y, z) relative to the moving system Oxyz is relative, relative to the stationary system O1x1y1z1 is absolute. The motion of the mobile system Oxyz relative to the stationary system O1x1y1z1 is a portable motion. 14 z x1 y1 z1 O1 xy M xyz O Basic equation of dynamics: Absolute acceleration of a point: Substitute the absolute acceleration of a point into the basic equation of dynamics: Transfer terms with translational and Coriolis acceleration to the right side: The transferred terms have the dimension of forces and are considered as the corresponding forces of inertia, equal: Then the relative motion of a point can be considered as absolute, if we add the translational and Coriolis forces of inertia to the acting forces: In the projections on the axis of the moving coordinate system, we have: rotation is uniform, then εe = 0: 2. Translational curvilinear motion: If the motion is rectilinear, then =: If the motion is rectilinear and uniform, then the moving system is inertial and the relative motion can be considered as absolute: motion (the principle of relativity of classical mechanics). Influence of the Earth's rotation on the balance of bodies - Let us assume that the body is in equilibrium on the surface of the Earth at an arbitrary latitude φ (parallel). The Earth rotates on its axis from west to east with an angular velocity: The radius of the Earth is about 6370 km. S R - full reaction of a non-smooth surface. G is the force of gravity of the Earth to the center. Ф - centrifugal force of inertia. Condition of relative equilibrium: The resultant of the forces of attraction and inertia is the force of gravity (weight): The magnitude of the force of gravity (weight) on the Earth's surface is equal to P = mg. The centrifugal force of inertia is a small fraction of the force of gravity: The deviation of the force of gravity from the direction of the force of gravity is also small: Thus, the influence of the rotation of the Earth on the balance of bodies is extremely small and is not taken into account in practical calculations. The maximum value of the force of inertia (at φ = 0 - at the equator) is only 0.00343 of the value of the force of gravity
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Lecture 5 (continued 5.2) 15 Influence of the Earth's rotation on the motion of bodies in the gravitational field of the Earth - Let's put the body falls on the Earth from a certain height H above the Earth's surface at latitude φ. Let's choose a moving frame of reference rigidly connected to the Earth, directing the x and y axes tangentially to the parallel and to the meridian: Thus, the force of gravity is identified with the force of gravity. In addition, we believe that the force of gravity is directed perpendicular to the surface of the Earth due to the smallness of its deflection, as discussed above. Coriolis acceleration is equal and directed parallel to the y-axis to the west. The Coriolis inertial force is equal to the opposite direction. Let's project the equation of relative motion on the axis: The solution of the first equation gives: Initial conditions: The solution of the third equation gives: Initial conditions: The third equation takes the form: Initial conditions: Its solution gives: The obtained solution shows that the body deviates to the east when falling. Let us calculate the magnitude of this deviation, for example, when falling from a height of 100 m. The time of the fall is found from the solution of the second equation: Thus, the influence of the Earth's rotation on the motion of bodies is extremely small for practical heights and speeds and is not taken into account in technical calculations. The solution to the second equation also implies the existence of a velocity along the y-axis, which should also cause and cause the corresponding acceleration and Coriolis inertial force. The influence of this speed and the force of inertia associated with it on the change in motion will be even less than the considered Coriolis inertial force associated with the vertical speed.
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Lecture 6 Dynamics of a mechanical system. A system of material points or a mechanical system - A set of material points or material ones united by general laws of interaction (the position or movement of each of the points or a body depends on the position and movement of all the others) A system of free points - the movement of which is not limited by any connections (for example, a planetary system , in which the planets are considered material points). A system of non-free points or a non-free mechanical system - the movement of material points or bodies is limited by the constraints imposed on the system (for example, a mechanism, a machine, etc.). 16 Forces acting on the system. In addition to the previously existing classification of forces (active and reactive forces), a new classification of forces is introduced: 1. External forces (e) - acting on points and bodies of the system from points or bodies that are not part of this system. 2. Internal forces (i) - forces of interaction between material points or bodies included in this system. One and the same force can be both external and internal force. It all depends on which mechanical system is being considered. For example: In the Sun, Earth and Moon system, all gravitational forces between them are internal. When considering the system the Earth and the Moon, the gravitational forces applied from the Sun are external: C З Л Based on the law of action and reaction, each internal force Fk corresponds to another internal force Fk ', equal in magnitude and opposite in direction. Two remarkable properties of internal forces follow from this: The main vector of all internal forces of the system is equal to zero: The main moment of all internal forces of the system relative to any center is equal to zero: Or in projections onto the coordinate axes: Remark. Although these equations are similar to the equilibrium equations, they are not, since internal forces are applied to various points or bodies of the system and can cause these points (bodies) to move relative to each other. It follows from these equations that internal forces do not affect the motion of the system considered as a whole. The center of mass of the system of material points. To describe the motion of the system as a whole, a geometric point is introduced, called the center of mass, the radius vector of which is determined by the expression, where M is the mass of the entire system: Or in projections onto the coordinate axes: The formulas for the center of mass are similar to the formulas for the center of gravity. However, the concept of the center of mass is more general since it is not related to gravitational forces or gravity forces.
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Lecture 6 (continued 6.2) 17 Theorem on the motion of the center of mass of the system - Consider a system of n material points. We divide the forces applied to each point into external and internal ones and replace them with the corresponding resultants Fke and Fki. Let us write down for each point the basic equation of dynamics: or Sum up these equations over all points: On the left side of the equation we introduce the masses under the sign of the derivative and replace the sum of the derivatives with the derivative of the sum: From the definition of the center of mass: Substitute into the resulting equation: After removing the mass of the system outside the sign of the derivative we get or: The product of the mass of the system and the acceleration of its center, the mass is equal to the main vector of external forces. In projections on the coordinate axes: The center of mass of the system moves like a material point with a mass equal to the mass of the entire system, to which all external forces acting on the system are applied. Consequences from the theorem on the motion of the center of mass of the system (conservation laws): 1. If in the time interval the main vector of the external forces of the system is equal to zero, Re = 0, then the velocity of the center of mass is constant, vC = const (the center of mass moves uniformly rectilinear - the law of conservation of motion center of mass). 2. If in the time interval the projection of the main vector of the external forces of the system on the x-axis is equal to zero, Rxe = 0, then the velocity of the center of mass along the x-axis is constant, vCx = const (the center of mass moves uniformly along the axis). Similar statements are true for the y and z axes. Example: Two people of masses m1 and m2 are in a boat of masses m3. At the initial moment of time, the boat with people was at rest. Determine the movement of the boat if a person weighing m2 has moved to the bow of the boat at a distance a. 3. If in the time interval the main vector of the external forces of the system is equal to zero, Re = 0, and at the initial moment the velocity of the center of mass is zero, vC = 0, then the radius vector of the center of mass remains constant, rC = const (the center of mass is at rest - the law of conservation of the position of the center of mass). 4. If in the time interval the projection of the main vector of the external forces of the system onto the x axis is zero, Rxe = 0, and at the initial moment the velocity of the center of mass along this axis is zero, vCx = 0, then the coordinate of the center of mass along the x axis remains constant, xC = const (the center of mass does not move along this axis). Similar statements are true for the y and z axes. 1. Object of movement (boat with people): 2. We discard connections (water): 3. Replace the connection with reaction: 4. Add active forces: 5. Write down the theorem about the center of mass: Project onto the x-axis: O Determine how far to change seats to a person of mass m1 so that the boat remains in place: The boat will move a distance l in the opposite direction.
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Lecture 7 Impulse of force - a measure of mechanical interaction, characterizing the transfer of mechanical motion from the side of forces acting on a point for a given period of time: 18 to a point of forces in the same period of time: Multiply by dt: We will integrate over a given time interval: The amount of movement of a point is a measure of mechanical movement, determined by a vector equal to the product of the mass of a point by the vector of its velocity: Theorem on the change in the momentum of the system - Consider the system n material points. We divide the forces applied to each point into external and internal ones and replace them with the corresponding resultants Fke and Fki. Let us write down for each point the basic equation of dynamics: or The amount of motion of the system of material points is the geometric sum of the amounts of movement of material points: By definition of the center of mass: The vector of the momentum of the system is equal to the product of the mass of the entire system by the velocity vector of the center of mass of the system. Then: In projections on the coordinate axes: The derivative of the vector of the momentum of the system with respect to time is equal to the main vector of the external forces of the system. Let's sum up these equations over all points: On the left side of the equation, we introduce the masses under the sign of the derivative and replace the sum of the derivatives with the derivative of the sum: From the definition of the momentum of the system: In projections onto the coordinate axes:
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Euler's theorem - Application of the theorem about the change in the momentum of the system to the motion of a continuous medium (water). 1. We select the volume of water in the curvilinear channel of the turbine as the object of motion: 2. We discard the constraints and replace their action with reactions (Rпов - resultant of surface forces) 3. Add active forces (Rпов - resultant of volumetric forces): 4. Write down the theorem about change in the amount of motion of the system: The amount of movement of water at times t0 and t1 is represented as the sum: Change in the amount of movement of water in the time interval: Change in the amount of movement of water for an infinitely small time interval dt:, where F1 F2 Taking the product of density, cross-sectional area and velocity for a second mass we obtain: Substituting the differential of the momentum of the system into the change theorem, we obtain: Consequences from the theorem on the change in the momentum of the system (conservation laws): 1. If in the time interval the main vector of the external forces of the system is equal to zero, Re = 0, then the quantity vector of motion is constant, Q = const is the law of conservation of the momentum of the system). 2. If in the time interval the projection of the main vector of the external forces of the system on the x axis is equal to zero, Rxe = 0, then the projection of the momentum of the system on the x axis is constant, Qx = const. Similar statements are true for the y and z axes. Lecture 7 (continued 7.2) Example: A grenade of mass M, flying with a speed of v, exploded into two parts. The speed of one of the fragments of mass m1 increased in the direction of motion to the value v1. Determine the speed of the second shard. 1. The object of movement (grenade): 2. The object is a free system, connections and their reactions are absent. 3. Add active forces: 4. Write down the theorem about the change in the momentum: Project on the axis: β Separate the variables and integrate: The right integral is practically zero, since explosion time t
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Lecture 7 (continued 7.3) 20 The angular momentum of a point or the angular momentum of motion relative to a certain center is a measure of mechanical motion, determined by a vector equal to the vector product of the radius vector of a material point by the vector of its momentum: The kinetic moment of a system of material points relative to a certain center is geometric the sum of the moments of the number of movements of all material points relative to the same center: In projections on the axis: In projections on the axis: The theorem on the change in the angular momentum of the system - Consider a system of n material points. We divide the forces applied to each point into external and internal ones and replace them with the corresponding resultants Fke and Fki. Let us write down for each point the basic equation of dynamics: or Sum these equations over all points: Replace the sum of derivatives by the derivative of the sum: The expression in brackets is the moment of moment of the system's momentum. From here: We multiply each of the equalities vector by the radius vector on the left: Let's see if it is possible to move the sign of the derivative outside the vector product: Thus, we got: The derivative of the angular momentum of the system relative to some center in time is equal to the main moment of the external forces of the system relative to the same center. In projections on the coordinate axes: The derivative of the angular momentum of the system relative to a certain axis in time is equal to the main moment of the external forces of the system relative to the same axis.
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Lecture 8 21 ■ Consequences from the theorem on the change in the angular momentum of the system (conservation laws): 1. If in the time interval the vector of the main moment of the external forces of the system with respect to some center is equal to zero, MOe = 0, then the vector of the angular momentum of the system relative to the same center constant, KO = const is the law of conservation of the angular momentum of the system). 2. If in the time interval the main moment of the external forces of the system relative to the x axis is equal to zero, Mxe = 0, then the angular momentum of the system relative to the x axis is constant, Kx = const. Similar statements are true for the y and z axes. 2. The moment of inertia of a rigid body about the axis: The moment of inertia of a material point about the axis is equal to the product of the mass of the point by the square of the distance of the point to the axis. The moment of inertia of a rigid body about an axis is equal to the sum of the products of the mass of each point by the square of the distance of this point to the axis. ■ Elements of the theory of moments of inertia - When a rigid body rotates, the measure of inertia (resistance to change in motion) is the moment of inertia about the axis of rotation. Let's consider the basic concepts of definition and methods of calculating the moments of inertia. 1. Moment of inertia of a material point about the axis: When passing from a discrete small mass to an infinitely small mass of a point, the limit of such a sum is determined by the integral: the axial moment of inertia of a rigid body. In addition to the axial moment of inertia of a rigid body, there are other types of moments of inertia: the centrifugal moment of inertia of a rigid body. polar moment of inertia of a rigid body. 3. The theorem on the moments of inertia of a rigid body about parallel axes - the formula for the transition to parallel axes: Moment of inertia about the original axis Static moments of inertia about the original axes Mass of the body Distance between the axes z1 and z2 Thus: If the axis z1 passes through the center of mass, then static moments are equal to zero:
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Lecture 8 (continued 8.2) 22 Moment of inertia of a homogeneous rod of constant cross-section about the axis: xz L Let's select the elementary volume dV = Adx at a distance x: x dx Elementary mass: To calculate the moment of inertia about the central axis (passing through the center of gravity), it is enough to change the position of the axis and set the limits of integration (-L / 2, L / 2). Here we will demonstrate the formula for the transition to parallel axes: zС 5. The moment of inertia of a homogeneous solid cylinder about the axis of symmetry: H dr r Let's select the elementary volume dV = 2πrdrH (thin cylinder of radius r): Elementary mass: Here we used the formula for the volume of a cylinder V = πR2H. To calculate the moment of inertia of a hollow (thick) cylinder, it is sufficient to set the limits of integration from R1 to R2 (R2> R1): 6. Moment of inertia of a thin cylinder about the axis of symmetry (t
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Lecture 8 (continued 8.3) 23 ■ The differential equation of rotation of a rigid body about an axis: Let us write a theorem about the change in the angular momentum of a rigid body rotating about a fixed axis: The kinetic moment of a rotating rigid body is: The moment of external forces about the axis of rotation is equal to the torque (reactions and force no moment gravity): Substitute the angular momentum and the torque in the theorem Example: Two people of the same weight G1 = G2 hang on a rope thrown over a solid block with weight G3 = G1 / 4. At some point, one of them began to climb the rope with a relative speed u. Determine the lifting speed of each of the people. 1. Select the object of motion (block with people): 2. Discard the connections (support device of the block): 3. Replace the connection with reactions (bearing): 4. Add active forces (gravity): 5. Write down the theorem about the change in the kinetic moment of the system with respect to the axes of rotation of the block: R Since the moment of external forces is equal to zero, the angular momentum must remain constant: At the initial moment of time t = 0, there was equilibrium and Kz0 = 0. After the beginning of the movement of one person relative to the rope, the whole system began to move, but the angular momentum system must remain equal to zero: Kz = 0. The kinetic moment of the system is the sum of the kinetic moments of both people and the block: Here v2 is the speed of the second person, equal to the speed of the cable, Example: Determine the period of small free oscillations of a homogeneous rod of mass M and length l, suspended by one end to the fixed axis of rotation. Or: In the case of small oscillations sinφ φ: Oscillation period: Moment of inertia of the bar:
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Lecture 8 (continued 8.4 - additional material) 24 ■ Elementary theory of a gyroscope: A gyroscope is a rigid body rotating around an axis of material symmetry, one of the points of which is motionless. A free gyroscope is fixed so that its center of mass remains stationary, and the axis of rotation passes through the center of mass and can take any position in space, i.e. the axis of rotation changes its position like the axis of its own rotation of the body during spherical motion. The main assumption of the approximate (elementary) theory of the gyroscope is that the vector of the angular momentum (angular momentum) of the rotor is assumed to be directed along its own axis of rotation. Thus, despite the fact that in the general case the rotor participates in three rotations, only the angular velocity of its own rotation ω = dφ / dt is taken into account. The reason for this is that in modern technology, the gyroscope rotor rotates at an angular velocity of the order of 5000-8000 rad / s (about 50,000-80000 rpm), while the other two angular velocities associated with the precession and nutation of its own axis of rotation tens of thousands of times less than this speed. The main property of a free gyroscope is that the rotor axis maintains a constant direction in space with respect to the inertial (stellar) frame of reference (demonstrated by the Foucault pendulum, which keeps the swing plane unchanged with respect to the stars, 1852). This follows from the law of conservation of the angular momentum relative to the center of mass of the rotor, provided that friction in the bearings of the rotor suspension axles, external and internal frames is neglected: Force action on the axis of a free gyroscope. In the case of the action of a force applied to the rotor axis, the moment of external forces relative to the center of mass is not equal to zero: force, and in the direction of the vector of the moment of this force, i.e. will pivot not about the x-axis (internal suspension), but about the y-axis (external suspension). When the force is terminated, the rotor axis will remain in the unchanged position corresponding to the last moment of time of the force, because from this moment in time, the moment of external forces again becomes equal to zero. In the case of a short-term action of force (impact), the gyroscope axis practically does not change its position. Thus, the rapid rotation of the rotor imparts to the gyroscope the ability to counteract random influences tending to change the position of the axis of rotation of the rotor, and under constant action of the force maintains the position of the plane perpendicular to the acting force in which the axis of the rotor lies. These properties are used in the operation of inertial navigation systems.
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- Statics
- Basic concepts of statics
- Types of forces
- Axioms of statics
- Connections and their reactions
- System of converging forces
- Methods for determining the resultant system of converging forces
- Equilibrium conditions for a system of converging forces
- Moment of force relative to the center as a vector
- Algebraic magnitude of the moment of force
- Properties of the moment of force about the center (point)
- The theory of pairs of forces
- Addition of two parallel forces directed in one direction
- The addition of two parallel forces directed in opposite directions
- Couples of forces
- Pair of forces theorems
- Equilibrium conditions for a system of pairs of forces
- Lever arm
- Arbitrary flat system of forces
- Cases of Reducing a Plane System of Forces to a Simpler Form
- Analytical Equilibrium Conditions
- Center of Parallel Forces. The center of gravity
- Center of Parallel Forces
- The center of gravity of a rigid body and its coordinates
- Center of gravity of volume, plane and line
- Methods for determining the position of the center of gravity
- Basics of strength calculations
- Tasks and methods of strength of materials
- Classification of loads
- Classification of structural elements
- Bar deformations
- Basic hypotheses and principles
- Internal forces. Section method
- Voltage
- Stretching and squeezing
- Mechanical characteristics of the material
- Allowable voltages
- Hardness of materials
- Plots of longitudinal forces and stresses
- Shift
- Geometric characteristics of sections
- Torsion
- Bend
- Differential Bending Constraints
- Flexural strength
- Normal voltages. Strength calculation
- Shear bending stresses
- Bending stiffness
- Elements of the general theory of the stress state
- Strength theories
- Torsion bend
- Kinematics
- Point kinematics
- Point trajectory
- Methods for specifying point movement
- Point speed
- Point acceleration
- Rigid body kinematics
- The translational motion of a rigid body
- Rotational motion of a rigid body
- Gear kinematics
- Plane-parallel movement of a rigid body
- Complex point movement
- Point kinematics
- Dynamics
- Basic laws of dynamics
- Point dynamics
- Differential equations of a free material point
- Two problems of point dynamics
- Rigid body dynamics
- Classification of forces acting on a mechanical system
- Differential equations of motion of a mechanical system
- General theorems of dynamics
- The theorem on the motion of the center of mass of a mechanical system
- Momentum Change Theorem
- The theorem on the change in the angular momentum
- The theorem on the change in kinetic energy
- Forces acting in machines
- Forces in engagement of a spur gear
- Friction in mechanisms and machines
- Sliding friction
- Rolling friction
- Efficiency
- Machine parts
- Mechanical transmission
- Types of mechanical transmissions
- Basic and derived parameters of mechanical transmissions
- Gear transmission
- Flexible link transmissions
- Shafts
- Purpose and classification
- Design calculation
- Check calculation of shafts
- Bearings
- Plain bearings
- Rolling bearings
- Connecting machine parts
- Types of detachable and one-piece connections
- Keyed connections
- Mechanical transmission
- Standardization of norms, interchangeability
- Tolerances and landings
- Unified System of Tolerances and Landings (ESDP)
- Geometric Tolerance and Position
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An example of calculating a spur gear
An example of calculating a spur gear. The choice of material, calculation of permissible stresses, calculation of contact and bending strength were performed.
An example of solving the problem of bending a beam
In the example, diagrams of shear forces and bending moments are constructed, a dangerous section is found and an I-beam is selected. The problem analyzes the construction of diagrams using differential dependencies, a comparative analysis of various cross-sections of the beam is carried out.
An example of solving the problem of shaft torsion
The task is to check the strength of a steel shaft for a given diameter, material and allowable stresses. During the solution, diagrams of torques, shear stresses and torsion angles are plotted. The dead weight of the shaft is not taken into account.
An example of solving the problem of tension-compression of a bar
The task is to check the strength of a steel bar at a given allowable stress. In the course of the solution, diagrams of longitudinal forces, normal stresses and displacements are plotted. The self-weight of the bar is not taken into account.
Application of the kinetic energy conservation theorem
An example of solving the problem on the application of the theorem on the conservation of kinetic energy of a mechanical system
Determination of the speed and acceleration of a point according to the given equations of motion
An example of solving a problem to determine the speed and acceleration of a point according to the given equations of motion
Determination of the velocities and accelerations of points of a rigid body during plane-parallel motion
An example of solving the problem of determining the velocities and accelerations of points of a rigid body during plane-parallel motion
Determination of forces in the bars of a flat truss
An example of solving the problem of determining the forces in the bars of a flat truss by the Ritter method and by the method of cutting nodes
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