Basic mechanics for dummies. Introduction. Course of lectures on theoretical mechanics. Dynamics Popular short notes on theoretical mechanics

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  • Statics
    • Basic concepts of statics
    • Force types
    • Axioms of statics
    • Connections and their reactions
    • Converging force system
      • Methods for determining the resultant system of converging forces
      • Equilibrium conditions for a system of converging forces
    • Moment of force about the center as a vector
      • Algebraic value of the moment of force
      • Properties of the moment of force about the center (point)
    • Theory of pairs of forces
      • Addition of two parallel forces in the same direction
      • Addition of two parallel forces directed in opposite directions
      • Power Pairs
      • Couple of forces theorems
      • Conditions for the equilibrium of a system of pairs of forces
    • Lever arm
    • Arbitrary plane system of forces
      • Cases of reducing a flat system of forces to a simpler form
      • Analytical equilibrium conditions
    • Center of Parallel Forces. Center of gravity
      • Center of Parallel Forces
      • Center of gravity solid body and its coordinates
      • Center of gravity of volume, planes and lines
      • Methods for determining the position of the center of gravity
  • Basics of Strength Racsets
    • Problems and methods of resistance of materials
    • Load classification
    • Classification of structural elements
    • Rod deformations
    • Main hypotheses and principles
    • Internal forces. Section method
    • Voltage
    • Tension and compression
    • Mechanical characteristics of the material
    • Permissible stresses
    • Material hardness
    • Plots of longitudinal forces and stresses
    • Shift
    • Geometric characteristics of sections
    • Torsion
    • bend
      • Differential dependencies in bending
      • Flexural strength
      • normal stresses. Strength calculation
      • Shear stresses in bending
      • Bending stiffness
    • Elements general theory stress state
    • Strength theories
    • Bending with twist
  • Kinematics
    • Point kinematics
      • Point trajectory
      • Methods for specifying the movement of a point
      • Point speed
      • point acceleration
    • Rigid Body Kinematics
      • Translational motion of a rigid body
      • Rotational motion of a rigid body
      • Kinematics of gear mechanisms
      • Plane-parallel motion of a rigid body
    • Complex point movement
  • 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
      • Theorem on the motion of the center of mass of a mechanical system
      • Theorem on the change in momentum
      • Theorem on the change in angular momentum
      • Kinetic energy change theorem
  • Forces acting in machines
    • Forces in engagement of a spur gear
    • Friction in mechanisms and machines
      • Sliding friction
      • rolling friction
    • Efficiency
  • Machine parts
    • Mechanical transmissions
      • Types of mechanical gears
      • Basic and derived parameters of mechanical gears
      • gears
      • Gears with flexible links
    • Shafts
      • Purpose and classification
      • Design calculation
      • Check calculation of shafts
    • Bearings
      • Plain bearings
      • Rolling bearings
    • Connection of machine parts
      • Types of detachable and permanent connections
      • Keyed connections
  • Standardization of norms, interchangeability
    • Tolerances and landings
    • one system tolerances and landings (ESDP)
    • Form and position deviation

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An example of the calculation of a spur gear
An example of the calculation of a spur gear. The choice of material, the calculation of allowable stresses, the calculation of contact and bending strength were carried out.


An example of solving the problem of beam bending
In the example, diagrams of transverse forces and bending moments are plotted, a dangerous section is found, and an I-beam is selected. In the problem, the construction of diagrams using differential dependencies was analyzed, comparative analysis different cross sections of the beam.


An example of solving the problem of shaft torsion
The task is to test the strength of a steel shaft for a given diameter, material and allowable stresses. During the solution, diagrams of torques, shear stresses and twist angles are built. Self weight of the shaft is not taken into account


An example of solving the problem of tension-compression of a rod
The task is to test the strength of a steel rod at given allowable stresses. During the solution, plots of longitudinal forces, normal stresses and displacements are built. Self weight of the bar is not taken into account


Application of the kinetic energy conservation theorem
An example of solving the problem of applying 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 the problem of determining the speed and acceleration of a point according to the given equations of motion


Determination of 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 Planar Truss Bars
An example of solving the problem of determining the forces in the bars of a flat truss by the Ritter method and the knot cutting method

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Course of lectures on theoretical mechanics Dynamics (I part) Bondarenko A.N. Moscow - 2007 Electronic training course written on the basis of lectures given by the author for students studying in the specialties of SZhD, PGS and SDM at NIIZhT and MIIT (1974-2006). Educational material corresponds to the calendar plans in the amount of three semesters. For the full implementation of animation effects during the 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 Railways (MIIT) Department of Theoretical Mechanics Scientific and Technical Center of Transport Technologies

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Contents Lecture 1. Introduction to dynamics. Laws and axioms of material point dynamics. 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. Solving the inverse problem of dynamics. General instructions for solving the inverse problem of dynamics. Examples of solving the inverse problem of dynamics. The motion of a body thrown at an angle to the horizon, without taking into account air resistance. Lecture 3. Rectilinear oscillations of a material point. The condition for the occurrence of oscillations. Classification of vibrations. Free vibrations without taking into account the forces of resistance. damped vibrations. Oscillation decrement. Lecture 4. Forced oscillations of a material point. Resonance. Influence of resistance to motion during forced vibrations. Lecture 5. Relative motion of a material point. Forces of inertia. Particular cases of movement for various types of portable movement. 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. Center of mass of the system. The theorem on the motion of the center of mass. Conservation laws. An example of solving the problem of using the theorem on the movement of the center of mass. Lecture 7. Impulse of force. The amount of movement. Theorem on the change in momentum. Conservation laws. Euler's theorem. An example of solving the problem on the use of the theorem on the change in momentum. moment of momentum. The theorem on changing the angular momentum. Lecture 8. Conservation laws. Elements of the theory of moments of inertia. Kinetic moment of a rigid body. Differential equation rotation of a rigid body. An example of solving the problem of using the theorem on changing the angular momentum of the system. Elementary theory of the gyroscope. Recommended literature 1. Yablonsky A.A. Course of theoretical mechanics. Part 2. M.: graduate School. 1977. 368 p. 2. Meshchersky I.V. Collection of problems in theoretical mechanics. M.: Science. 1986 416 p. 3. Collection of tasks for term papers/ Ed. A.A. Yablonsky. M.: Higher school. 1985. 366 p. 4. Bondarenko A.N. “Theoretical mechanics in examples and tasks. Dynamics” ( electronic manual www.miit.ru/institut/ipss/faculties/trm/main.htm), 2004

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Lecture 1 Dynamics is a section of theoretical mechanics that studies mechanical motion from the very beginning. common point vision. 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 motion of a material point, taking into account the forces that cause 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 movement. - there is an absolute time (does not depend on matter and its movement). It follows from this: - there is an absolutely immobile frame of reference. - time does not depend on 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 scope, since the mechanical systems considered in applied sciences do not have such large masses and speeds of movement, for which it is necessary to take into account their influence on the geometry of space, time, movement, as is done in relativistic mechanics (the theory of relativity) ■ The basic laws of dynamics - first discovered by Galileo and formulated by Newton form the basis of all methods for describing and analyzing the movement of mechanical systems and their dynamic interaction action under the influence of various forces. ■ Law of inertia (Galileo-Newton law) - An isolated material point of a body retains its state of rest or uniform rectilinear motion until the forces applied cause it to change that state. This implies the equivalence of the state of rest and motion by inertia (the law of relativity of Galileo). 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 (a measure of inertia), measured in kg, numerically equal to weight divided by the acceleration of gravity: F is the acting force, measured in N (1 N imparts an acceleration of 1 m / s2 to a point of mass 1 kg, 1 N \u003d 1 / 9.81 kg-s). ■ Dynamics of a mechanical system - studies the movement of a set of material points and solid bodies, united by the general laws of interaction, taking into account the forces that cause this movement. ■ Analytical mechanics - studies the motion of non-free mechanical systems using general analytical methods. one

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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 point motion in coordinate form. This result can be obtained by formal projection of the vector differential equation (1). After grouping, the vector relation is decomposed into three scalar equations: In coordinate form: We use the relationship of the radius-vector with coordinates and the force vector with projections: differential equation of motion on natural (moving) coordinate axes: or: - natural equations of motion of a point. ■ Basic equation of dynamics: - corresponds to the vector way of specifying the movement 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 (Newton's III law) - Every action corresponds to an equal 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 are given. It is required to find motion parameters (motion equations, motion trajectory). 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 release 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 of the first problem is connected 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 difficult than the direct problem. The solution of the direct problem of dynamics - let's look at examples: Example 1. A cabin with a weight G of an elevator is lifted by a cable with an acceleration a . Determine cable tension. 1. Select an object (the elevator car moves forward and can be considered as a material point). 2. We discard the connection (cable) and replace it with the reaction R. 3. Compile the basic equation of dynamics: Determine the reaction of the cable: Determine the cable tension: With a uniform movement of the cab ay = 0 and the cable tension 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 free fall: ay = -g. 3 4. We project the basic equation of dynamics onto the y axis: y Example 2. A point of mass m moves along a horizontal surface (the 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 the reaction N. 3. Add an unknown force F to the system of forces. 4. Compose the basic equation of dynamics: 5. Project the basic equation of dynamics onto axes x,y: Determine force projections: Force modulus: Direction cosines: Thus, the magnitude of the force is proportional to the distance of the point to the center of coordinates and is directed towards the center along the line connecting the point to the center. The trajectory of the point movement is an ellipse centered at the origin: O r Lecture 1 (continued - 1.3)

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Lecture 1 (continuation 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 with a certain speed. The angle of deviation of the cable from the vertical is equal to. Determine the tension of the cable and the speed of the load. 1. Select an object (cargo). 2. Discard the connection (rope) and replace it with the reaction R. 3. Compose the basic equation of dynamics: From the third equation, determine the reaction of the cable: Determine the tension of the cable: Substitute the value of the reaction of the cable, normal acceleration into the second equation and determine the speed of the load: 4. Project the main equation axle dynamics,n,b: Example 4: A car of weight G moves on a convex bridge (radius of curvature is R) with 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 connection (rough surface) and replace it with the reactions N and the friction force Ffr. 3. We compose the basic equation of dynamics: 4. We project the basic equation of dynamics onto the n axis: 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 substituting 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 movements determined by the initial conditions. The initial coordinates take into account the initial position of the point. The initial velocity, given by the projections, takes into account the influence on its movement along the considered section of the trajectory of the forces that acted on the point before arriving at this section, i.e. initial kinematic state. Solution of the inverse problem of dynamics - In the general case of the movement of a point, the forces acting on the 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: Example 1 of the solution inverse problem: A free material point of mass m moves under the action of a force F, which is 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. Compose the basic equation of dynamics: 3. Lower the order of the derivative: 2. Choose a Cartesian reference system, directing the x axis along the direction of the force and project the basic equation of dynamics onto this axis: or x y z 4. Separate the variables: 5. Calculate the integrals of both parts of the equation : 6. Let's represent the velocity projection as a derivative of the coordinate with respect to time: 8. Calculate the integrals of both parts 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 obtain the equation of uniformly variable motion (along the x axis): 5

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General instructions for solving direct and inverse problems. Solution procedure: 1. Compilation of the differential equation of motion: 1.1. Choose a coordinate system - rectangular (fixed) with an unknown trajectory of movement, natural (moving) with a known trajectory, for example, a circle or a straight line. In the latter case, one rectilinear coordinate can be used. The reference point should be combined 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 fluctuates. 6 1.2. Draw a point at a position corresponding to an arbitrary moment in time (for t > 0) so that the coordinates are positive (s > 0, x > 0). We also assume that the velocity projection in this position is also positive. In the case of oscillations, the velocity projection changes sign, for example, when 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. The implementation of this recommendation is important in the future when working with resistance forces that depend on speed. 1.3. Release the material point from bonds, replace their action with reactions, add active forces. 1.4. Write the basic law of dynamics in vector form, project onto the selected axes, express the given or reactive forces in terms of the variables time, coordinates or speeds, if they depend on them. 2. Solution of differential equations: 2.1. Reduce the derivative if the equation is not reduced to the canonical (standard) form. for example: or 2.2. Separate variables, for example: or 2.4. Calculate not definite integrals in the left and right parts 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 separate the variables. Comment. Instead of calculating indefinite integrals it is possible to calculate definite integrals with a variable upper limit. The lower limits represent the initial values ​​of the variables (initial conditions). Then there is no need to separately find the constant, 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 time derivative of the coordinate, for example, and repeat steps 2.2 -2.4 Note. If the equation is reduced to a canonical form that has a standard solution, then this is turnkey solution and is used. Integration constants are still found from the initial conditions. See, for example, oscillations (lecture 4, p. 8). Lecture 2 (continuation 2.2)

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Lecture 2 (continuation 2.3) Example 2 of solving the inverse problem: Force depends on time. A load of weight P begins to move along a smooth horizontal surface under the action of a force F, the magnitude of which is proportional to time (F = kt). Determine the distance traveled by the load in time t. 3. We compose the main equation of dynamics: 5. We lower the order of the derivative: 4. We project the main equation of dynamics onto the x-axis: or 7 6. We separate the variables: 7. We calculate the integrals from both parts of the equation: 9. We represent the projection of the velocity as the derivative of the coordinate with respect to time: 10. Calculate the integrals of both parts 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 the reference system (Cartesian coordinates) so that the body has a positive coordinate: 2. We take the object of motion as a material point (the body moves forward), release it from the connection (reference plane) and replace it with the 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 upwards from the Earth's surface with a speed v0. The force of gravity of the Earth is inversely proportional to the square of the distance from the 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 reference system (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 onto 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. Change the variable: 6. Separate the variables: 7. Calculate the integrals of both sides of the equation: 8. Substitute the limits: As a result, we obtain an expression for the speed as a function of the y coordinate: Maximum height flight can be found by equating the speed to zero: The maximum flight altitude when the denominator turns to zero: From here, when setting the radius of the Earth and the acceleration of free fall, 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 v0. The resistance of water to the movement of the ship is proportional to the speed. Determine the time it takes the ship's speed to drop by half after turning off the engine, as well as the distance traveled by the ship to a complete stop. 8 1. We choose a reference system (Cartesian coordinates) so that the body has a positive coordinate: 2. We take the object of motion as a material point (the ship moves forward), free it from bonds (water) and replace it with a reaction (buoyant force - Archimedes force), and also the force of resistance to movement. 3. Add active force (gravity). 4. We compose the main equation of dynamics: 5. We project the main equation of dynamics onto the x-axis: or 6. We lower the order of the derivative: 7. We separate the variables: 8. We calculate the integrals from both parts of the equation: 9. We substitute the limits: An expression is obtained that relates the speed and time t, from which you can determine the time of movement: The time of movement, during which the speed will fall by half: It is interesting to note that when the speed approaches zero, the time of movement tends to infinity, i.e. final velocity cannot be zero. Why not "perpetual motion"? However, in this case, the distance traveled to the stop is a finite value. To determine the distance traveled, we turn to the expression obtained after lowering the order of the derivative, and make a change of variable: After integration and substitution of limits, we obtain: Distance traveled to a stop: ■ 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 - The oscillatory movement 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, the equilibrium position is stable No restoring force, the equilibrium position is unstable No restoring force, the equilibrium position is indifferent It is always directed towards the equilibrium position, the value is directly proportional to the linear elongation (shortening) of the spring, which is equal to the deviation of the body from the equilibrium position: c is the spring stiffness coefficient, 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 oscillations taking into account the resistance of the medium (damped oscillations). 3. Forced vibrations. 4. Forced oscillations taking into account the resistance of the medium. ■ Free oscillations - occur under the action of only a restoring force. Let's write down the basic law of dynamics: Let's choose a coordinate system centered at the equilibrium position (point O) and project the equation onto the x-axis: Let's bring the resulting equation to the standard (canonical) form: This equation is a homogeneous linear differential equation of the second order, the form of the solution of which is determined by the roots of the characteristic equation, obtained using the universal substitution: The roots of the characteristic equation are imaginary and equal: Common decision differential equation has the form: Point speed: Initial conditions: Let's define constants: So, the equation of free oscillations has the form: The equation can be represented by a one-term expression: where a is the amplitude, is the initial phase. The new constants a and - are related to the constants C1 and C2 by the relations: Define a and: The reason for the occurrence of free oscillations is the initial displacement x0 and/or starting speed v0.

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10 Lecture 3 (continuation 3.2) Damped oscillations of a material point - The oscillatory movement of a material point occurs in the presence of a restoring force and a force of resistance to movement. The dependence of the force of resistance to movement on displacement or speed is determined by the physical nature of the medium or connection that impedes movement. The simplest dependence is a linear dependence on speed (viscous resistance): - viscosity coefficient x y O from the values ​​of the roots: 1. n< k – случай малого вязкого сопротивления: - корни комплексные, различные. или x = ae-nt x = -ae-nt Частота damped oscillations: Period: T* Oscillation decrement: ai ai+1 Logarithmic oscillation decrement: Oscillation decays very quickly. The main influence of the viscous resistance force is a decrease in the oscillation amplitude over time. 2. n > k - the case of high viscous resistance: - the roots are real, different. or - these functions are aperiodic: 3. n = k: - roots are real, multiple. these functions are also aperiodic:

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Lecture 3 (continuation 3.3) Classification of solutions of free oscillations. Spring connections. equivalent hardness. y y 11 Diff. Equation Character. Equation Roots char. equation Solving 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 acts, called the perturbing force. The perturbing force can have a different nature. For example, in a particular case, the inertial effect of an unbalanced mass m1 of a rotating rotor causes harmonically changing force projections: The main equation of dynamics: The projection of the equation of dynamics on the axis: Let's 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 a particular solution of the inhomogeneous equation: We select the particular solution in the form of the right side: The resulting 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) vibrations. 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 the oscillations is opposite to the phase of the disturbing force:

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Lecture 4 (continuation 4.2) 13 Dynamic coefficient - the ratio of the amplitude of forced oscillations to the static deviation of a point under the action of a constant force H = const: The amplitude of forced oscillations: The static deviation 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 oscillations) dynamic coefficient: Resonance - occurs when the frequency of forced oscillations coincides with the frequency of natural oscillations (p = k). This most often occurs when starting and stopping the rotation of poorly balanced rotors mounted on elastic suspensions. The differential equation of oscillations with equal frequencies: A particular solution in the form of the right side cannot be taken, because a linearly dependent solution will be obtained (see the 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. Influence of resistance to motion 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, p. 11) depending on the ratio of n and k (see). We take a particular solution in the form and calculate the derivatives: Substitute in the differential equation: Equating the coefficients at the same trigonometric functions we obtain a system of equations: By raising both equations to a power and adding them, 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 oscillations, taking into account the resistance to movement, for example, at 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 - Let's assume that the moving (non-inertial) coordinate system Oxyz moves according to some law relative to the fixed (inertial) coordinate system O1x1y1z1. The motion of a material point M (x, y, z) relative to the mobile system Oxyz is relative, relative to the motionless system O1x1y1z1 is absolute. The motion of the mobile system Oxyz relative to the fixed system O1x1y1z1 is a portable motion. 14 z x1 y1 z1 O1 x y M x y z O absolute acceleration points in the main equation of dynamics: We transfer the terms with translational and Coriolis acceleration to the right side: The transferred terms have the dimension of forces and are considered as the corresponding inertia forces equal to: Then the relative motion of the point can be considered as absolute if we add the translational and Coriolis forces of inertia : In projections on the axes of the moving coordinate system, we have: Particular cases of relative motion of a point for various types of portable motion: 1. Rotation around a fixed axis: If the 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: No mechanical phenomena can detect a rectilinear uniform motion(principle of relativity of classical mechanics). Influence of the Earth's rotation on the equilibrium of bodies - Let's assume that the body is in equilibrium on the Earth's surface at an arbitrary latitude φ (parallels). The Earth rotates around its axis from west to east with an angular velocity: The radius of the Earth is about 6370 km. S R- full reaction uneven surface. G - force of attraction of the Earth to the center. Ф - centrifugal force of inertia. Relative equilibrium condition: 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 surface of the Earth is 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 attraction is also small: Thus, the influence of the Earth's rotation on the balance of bodies is extremely small and is not taken into account in practical calculations. The maximum value of the inertial force (at φ = 0 - at the equator) is only 0.00343 of the value of gravity

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Lecture 5 (continuation 5.2) 15 Influence of the Earth's rotation on the movement of bodies in the Earth's gravitational field - Suppose a body falls to 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 with the Earth, directing the x, y axes tangentially to the parallel and to the meridian: Relative motion equation: Here, the smallness of the centrifugal force of inertia compared to gravity is taken into account. Thus, the force of gravity is identified with the force of gravity. In addition, we assume that gravity is directed perpendicular to the Earth's surface due to the smallness of its deflection, as discussed above. The Coriolis acceleration is equal to and directed parallel to the y-axis to the west. The Coriolis inertia force is directed in the opposite direction. We 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 resulting solution shows that the body deviates to the east when it falls. Let us calculate the value of this deviation, for example, when falling from a height of 100 m. We find the fall time from the solution of the second equation: Thus, the influence of the Earth's rotation on the movement of bodies is extremely small for practical heights and speeds and is not taken into account in technical calculations. The solution of the second equation also implies the existence of a velocity along the y-axis, which should also cause and causes the corresponding acceleration and the Coriolis inertia force. The influence of this speed and the inertia force associated with it on the change in motion will be even less than the considered Coriolis inertia force associated with the vertical speed.

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Lecture 6 Dynamics of a mechanical system. System of material points or mechanical system - A set of material points or those material points 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) The 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 as 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 the given system. The same force can be both external and internal force. It all depends on which mechanical system is considered. For example: In the system of the Sun, Earth and Moon, all gravitational forces between them are internal. When considering the Earth and Moon system, the gravitational forces applied from the side of the Sun are external: C Z L Based on the law of action and reaction, each internal force Fk corresponds to another internal force Fk’, equal in absolute value 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: Note. Although these equations are similar to 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 a system considered as a whole. The center of mass of the system of material points. To describe the movement 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 those for the center of gravity. However, the concept of the center of mass is more general, since it is not related to the forces of gravity or the forces of gravity.

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Lecture 6 (continuation 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's write down for each point the basic equation of dynamics: or Let's sum these equations over all points: On the left side of the equation, we will 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: we obtain or: The product of the mass of the system and the acceleration of its center mass is equal to the main vector of external forces. In projections on the coordinate axes: The center of mass of the system moves as 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 speed of the center of mass is constant, vC = const (the center of mass moves uniformly rectilinearly - 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 mass m3. At the initial moment of time, the boat with people was at rest. Determine the displacement of the boat if a person of mass m2 moved to the bow of the boat at a distance a. 3. If in the time interval the main vector of external forces of the system is equal to zero, Re = 0, and at the initial moment the velocity of the center of mass is equal to zero, vC = 0, then the radius vector of the center of mass remains constant, rC = const (the center of mass is at rest is 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 on the x axis is equal to 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 motion (boat with people): 2. Discard connections (water): 3. Replace 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 you need to transfer to a person of mass m1, so that the boat stays in place: The boat will move a distance l in the opposite direction.

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Lecture 7 The impulse of force is a measure of mechanical interaction that characterizes the transfer of mechanical motion from the forces acting on the point for a given period of time: 18 In projections onto the coordinate axes: In the case of a constant force: In projections onto the coordinate axes: to the force point in the same time interval: Multiply by dt: Integrate over a given time interval: The amount of movement of the point is a measure of mechanical movement, determined by a vector equal to the product of the mass of the point and the vector of its velocity: Theorem on the change in the amount of movement 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's write for each point the basic equation of dynamics: or The momentum of the system of material points is the geometric sum of the quantities of motion 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 and the velocity vector of the center of mass of the system. Then: In projections onto the coordinate axes: The time derivative of the momentum vector of the system is equal to the main vector of the system's external forces. Let's sum 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 on the change in the momentum of a system to the movement of a continuous medium (water). 1. We select as the object of movement the volume of water located in the curvilinear channel of the turbine: 2. We discard the bonds and replace their action with reactions (Rpov - the resultant of surface forces) 3. Add active forces (Rb - the resultant of body forces): 4. Write down the theorem on change in the momentum of the system: The momentum of water at times t0 and t1 is represented as sums: Change in the momentum of water in the time interval : Change in the momentum of water over an infinitesimal time interval dt: , where F1 F2 Taking the product of density, cross-sectional area and velocity per 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 motion is constant, Q = const is the law of conservation of 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 (continuation of 7.2) Example: A grenade of mass M, flying at a speed 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 fragment. 1. The object of movement (grenade): 2. The object is a free system, there are no connections and their reactions. 3. Add active forces: 4. Write down the theorem on the change in momentum: Project onto the axis: β Divide the variables and integrate: The right integral is almost zero, because explosion time t

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Lecture 7 (continuation 7.3) 20 The angular momentum of a point or the kinetic moment 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 and 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 momentum of all material points relative to the same center: In projections on the axis: In projections on the axis: Theorem on the change in the moment of momentum of the system - Let's 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's write for each point the basic equation of dynamics: or Let's sum these equations over all points: Let's replace the sum of derivatives with the derivative of the sum: The expression in parentheses is the moment of momentum of the system. From here: We vectorially multiply each of the equalities by the radius-vector on the left: Let's see if it is possible to take the sign of the derivative beyond the limits of the vector product: Thus, we got: center. In projections on the coordinate axes: The derivative of the moment of momentum of the system relative to some 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 relative to a certain center is equal to zero, MOe = 0, then the vector of the angular momentum of the system relative to the same center is constant, KO = const is the law of conservation of 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. Moment of inertia of a rigid body about an axis: The moment of inertia of a material point about an axis is equal to the product of the mass of the point and 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 and the square of the distance of this point from the axis. ■ Elements of the theory of moments of inertia - With the rotational motion of a rigid body, the measure of inertia (resistance to change in motion) is the moment of inertia about the axis of rotation. Consider the basic concepts of the definition and methods for calculating the moments of inertia. 1. Moment of inertia of a material point about the axis: In the transition from a discrete small mass to an infinitely small mass of a point, the limit of such a sum is determined by the integral: 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 about the moments of inertia of a rigid body about parallel axes - the formula for the transition to parallel axes: Moment of inertia about the reference axis Static moments of inertia about the reference axes Body mass moments are zero:

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Lecture 8 (continuation 8.2) 22 Moment of inertia of a uniform rod of constant section about the axis: x z L 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 location of the axis and set the integration limits (-L/2, L/2). Here we 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 us single out the elementary volume dV = 2πrdrH (thin cylinder of radius r): Elementary mass: Here we use the cylinder volume formula V=πR2H. To calculate the moment of inertia of a hollow (thick) cylinder, it is enough to set the integration limits from R1 to R2 (R2> R1): 6. The moment of inertia of a thin cylinder about the axis of symmetry (t

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Lecture 8 (continuation 8.3) 23 ■ Differential equation of rotation of a rigid body about an axis: Let's write a theorem about changing the angular momentum of a rigid body rotating around a fixed axis: The momentum of a rotating rigid body is: The moment of external forces about the axis of rotation is equal to the torque (reactions and force do not create gravity moments): We substitute the kinetic moment and the torque into the theorem Example: Two people of the same weight G1 = G2 hang on a rope thrown over a solid block with a 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 person. 1. We select the object of motion (block with people): 2. We discard the connections (the support device of the block): 3. We replace the connection with reactions (bearing): 4. Add active forces (gravity): 5. Write down the theorem on the change in the kinetic moment of the system relative to axis of rotation of the block: R Since the moment of external forces is equal to zero, the kinetic moment must remain constant: At the initial moment of time t = 0, there was equilibrium and Kz0 = 0. After the start of the movement of one person relative to the rope, the whole system began to move, but the kinetic moment of the system must remain equal to zero: Kz = 0. The angular momentum of the system is the sum of the angular momentums of both people and the block: Here v2 is the speed of the second person, equal to the speed of the cable end to a fixed axis of rotation. Or: In the case of small oscillations sinφ φ: Period of oscillation: Moment of inertia of the rod:

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Lecture 8 (continuation 8.4 - additional material) 24 ■ Elementary theory of the gyroscope: A gyroscope is a rigid body rotating around the axis of material symmetry, one of the points of which is fixed. A free gyroscope is fixed in such a way 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 the body's own rotation during spherical motion. The main assumption of the approximate (elementary) theory of the gyroscope is that the momentum vector (kinetic moment) of the rotor is considered 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 angular velocity own rotation ω = dφ/dt. The reason for this is that in modern technology the gyroscope rotor rotates with an angular velocity of the order of 5000-8000 rad / s (about 50000-80000 rpm), while the other two angular velocities associated with precession and nutation of their 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) reference system (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 kinetic moment relative to the center of mass of the rotor, provided that the friction in the bearings of the rotor suspension axes, the outer and inner frame is neglected: Force action on the axis of a free gyroscope. In the case of a force applied to the rotor axis, the moment of external forces relative to the center of mass is not equal to zero: ω ω С force, and towards the vector of the moment of this force, i.e. will rotate not about the x-axis (internal suspension), but about the y-axis (external suspension). Upon termination of the force, the rotor axis will remain in the same position, corresponding to the last time of the force, because from this point in time, the moment of external forces again becomes equal to zero. In the case of a short-term action of force (impact), the axis of the gyroscope practically does not change its position. Thus, the rapid rotation of the rotor gives the gyroscope the ability to counteract random influences that seek to change the position of the axis of rotation of the rotor, and with a constant action of the force, it maintains the position of the plane perpendicular to the acting force in which the axis of the rotor lies. These properties are used in inertial systems navigation.

Theoretical mechanics- This is a branch of mechanics, which sets out the basic laws of mechanical motion and mechanical interaction of material bodies.

Theoretical mechanics is a science in which the movements of bodies over time (mechanical movements) are studied. It serves as the basis for other sections of mechanics (the theory of elasticity, resistance of materials, the theory of plasticity, the theory of mechanisms and machines, hydroaerodynamics) and many technical disciplines.

mechanical movement- this is a change over time in the relative position in space of material bodies.

Mechanical interaction- this is such an interaction, as a result of which the mechanical movement changes or the relative position of body parts changes.

Rigid body statics

Statics- This is a branch of theoretical mechanics, which deals with the problems of equilibrium of solid bodies and the transformation of one system of forces into another, equivalent to it.

    Basic concepts and laws of statics
  • Absolutely rigid body(solid body, body) is a material body, the distance between any points in which does not change.
  • Material point is a body whose dimensions, according to the conditions of the problem, can be neglected.
  • loose body is a body, on the movement of which no restrictions are imposed.
  • Non-free (bound) body is a body whose movement is restricted.
  • Connections- these are bodies that prevent the movement of the object under consideration (a body or a system of bodies).
  • Communication reaction is a force that characterizes the action of a bond on a rigid body. If we consider the force with which a rigid body acts on a bond as an action, then the reaction of the bond is a counteraction. In this case, the force - action is applied to the connection, and the reaction of the connection is applied to the solid body.
  • mechanical system is a set of interconnected bodies or material points.
  • Solid can be considered as a mechanical system, the positions and distance between the points of which do not change.
  • Strength is a vector quantity characterizing the mechanical action of one material body on another.
    Force as a vector is characterized by the point of application, the direction of action and the absolute value. The unit of measure for the modulus of force is Newton.
  • line of force is the straight line along which the force vector is directed.
  • Concentrated Power is the force applied at one point.
  • Distributed forces (distributed load)- these are forces acting on all points of the volume, surface or length of the body.
    The distributed load is given by the force acting per unit volume (surface, length).
    The dimension of the distributed load is N / m 3 (N / m 2, N / m).
  • External force is a force acting from a body that does not belong to the considered mechanical system.
  • inner strength is a force acting on a material point of a mechanical system from another material point belonging to the system under consideration.
  • Force system is the totality of forces acting on a mechanical system.
  • Flat system of forces is a system of forces whose lines of action lie in the same plane.
  • Spatial system of forces is a system of forces whose lines of action do not lie in the same plane.
  • Converging force system is a system of forces whose lines of action intersect at one point.
  • Arbitrary system of forces is a system of forces whose lines of action do not intersect at one point.
  • Equivalent systems of forces- these are systems of forces, the replacement of which one for another does not change the mechanical state of the body.
    Accepted designation: .
  • Equilibrium A state in which a body remains stationary or moves uniformly in a straight line under the action of forces.
  • Balanced system of forces- this is a system of forces that, when applied to a free solid body, does not change its mechanical state (does not unbalance it).
    .
  • resultant force is a force whose action on a body is equivalent to the action of a system of forces.
    .
  • Moment of power is a value that characterizes the rotational ability of the force.
  • Power couple is a system of two parallel equal in absolute value oppositely directed forces.
    Accepted designation: .
    Under the action of a couple of forces, the body will perform a rotational motion.
  • Projection of Force on the Axis- this is a segment enclosed between perpendiculars drawn from the beginning and end of the force vector to this axis.
    The projection is positive if the direction of the segment coincides with the positive direction of the axis.
  • Projection of Force on a Plane is a vector on a plane enclosed between the perpendiculars drawn from the beginning and end of the force vector to this plane.
  • Law 1 (law of inertia). An isolated material point is at rest or moves uniformly and rectilinearly.
    The uniform and rectilinear motion of a material point is a motion by inertia. The state of equilibrium of a material point and a rigid body is understood not only as a state of rest, but also as a movement by inertia. For a rigid body, there are various types of inertia motion, for example, uniform rotation of a rigid body around a fixed axis.
  • Law 2. A rigid body is in equilibrium under the action of two forces only if these forces are equal in magnitude and directed in opposite directions along common line actions.
    These two forces are called balanced.
    In general, forces are said to be balanced if the rigid body to which these forces are applied is at rest.
  • Law 3. Without violating the state (the word "state" here means the state of motion or rest) of a rigid body, one can add and discard balancing forces.
    Consequence. Without disturbing the state of a rigid body, the force can be transferred along its line of action to any point of the body.
    Two systems of forces are called equivalent if one of them can be replaced by another without disturbing the state of the rigid body.
  • Law 4. The resultant of two forces applied at one point is applied at the same point, is equal in absolute value to the diagonal of the parallelogram built on these forces, and is directed along this
    diagonals.
    The modulus of the resultant is:
  • Law 5 (law of equality of action and reaction). The forces with which two bodies act on each other are equal in magnitude and directed in opposite directions along one straight line.
    It should be borne in mind that action- force applied to the body B, and opposition- force applied to the body BUT, are not balanced, since they are attached to different bodies.
  • Law 6 (the law of hardening). The equilibrium of a non-solid body is not disturbed when it solidifies.
    It should not be forgotten that the equilibrium conditions, which are necessary and sufficient for a rigid body, are necessary but not sufficient for the corresponding non-rigid body.
  • Law 7 (the law of release from bonds). A non-free solid body can be considered as free if it is mentally freed from bonds, replacing the action of bonds with the corresponding reactions of bonds.
    Connections and their reactions
  • Smooth surface restricts movement along the normal to the support surface. The reaction is directed perpendicular to the surface.
  • Articulated movable support limits the movement of the body along the normal to the reference plane. The reaction is directed along the normal to the support surface.
  • Articulated fixed support counteracts any movement in a plane perpendicular to the axis of rotation.
  • Articulated weightless rod counteracts the movement of the body along the line of the rod. The reaction will be directed along the line of the rod.
  • Blind termination counteracts any movement and rotation in the plane. Its action can be replaced by a force presented in the form of two components and a pair of forces with a moment.

Kinematics

Kinematics- a branch of theoretical mechanics that deals with general geometric properties mechanical movement as a process occurring in space and time. Moving objects are considered as geometric points or geometric bodies.

    Basic concepts of kinematics
  • The law of motion of a point (body) is the dependence of the position of a point (body) in space on time.
  • Point trajectory is the locus of the positions of a point in space during its movement.
  • Point (body) speed- this is a characteristic of the change in time of the position of a point (body) in space.
  • Point (body) acceleration- this is a characteristic of the change in time of the speed of a point (body).
    Determination of the kinematic characteristics of a point
  • Point trajectory
    In the vector reference system, the trajectory is described by the expression: .
    In the coordinate reference system, the trajectory is determined according to the law of point motion and is described by the expressions z = f(x,y) in space, or y = f(x)- in the plane.
    In a natural reference system, the trajectory is predetermined.
  • Determining the speed of a point in a vector coordinate system
    When specifying the movement of a point in a vector coordinate system, the ratio of movement to the time interval is called the average value of the speed in this time interval: .
    Taking the time interval as an infinitesimal value, the value of the speed is obtained in this moment time (instantaneous speed value): .
    The average velocity vector is directed along the vector in the direction of the point movement, vector instantaneous speed directed tangentially to the trajectory in the direction of point movement.
    Conclusion: the speed of a point is a vector quantity equal to the derivative of the law of motion with respect to time.
    Derivative property: the time derivative of any value determines the rate of change of this value.
  • Determining the speed of a point in a coordinate reference system
    Rate of change of point coordinates:
    .
    The module of the full speed of a point with a rectangular coordinate system will be equal to:
    .
    The direction of the velocity vector is determined by the cosines of the steering angles:
    ,
    where are the angles between the velocity vector and the coordinate axes.
  • Determining the speed of a point in a natural reference system
    The speed of a point in a natural reference system is defined as a derivative of the law of motion of a point: .
    According to the previous conclusions, the velocity vector is directed tangentially to the trajectory in the direction of the point movement and in the axes is determined by only one projection .
    Rigid Body Kinematics
  • In the kinematics of rigid bodies, two main problems are solved:
    1) task of movement and determination of the kinematic characteristics of the body as a whole;
    2) determination of the kinematic characteristics of the points of the body.
  • Translational motion of a rigid body
    Translational motion is a motion in which a straight line drawn through two points of the body remains parallel to its original position.
    Theorem: in translational motion, all points of the body move along the same trajectories and at each moment of time have the same speed and acceleration in absolute value and direction.
    Conclusion: the translational motion of a rigid body is determined by the motion of any of its points, and therefore, the task and study of its motion is reduced to the kinematics of a point.
  • Rotational motion of a rigid body around a fixed axis
    The rotational motion of a rigid body around a fixed axis is the motion of a rigid body in which two points belonging to the body remain motionless during the entire time of movement.
    The position of the body is determined by the angle of rotation. The unit of measurement for an angle is radians. (A radian is the central angle of a circle whose arc length is equal to the radius, the full angle of the circle contains 2π radian.)
    Law rotary motion body around a fixed axis.
    The angular velocity and angular acceleration of the body will be determined by the differentiation method:
    — angular velocity, rad/s;
    — angular acceleration, rad/s².
    If we cut the body by a plane perpendicular to the axis, choose a point on the axis of rotation FROM and an arbitrary point M, then the point M will describe around the point FROM radius circle R. During dt there is an elementary rotation through the angle , while the point M will move along the trajectory for a distance .
    Linear speed module:
    .
    point acceleration M with a known trajectory is determined by its components:
    ,
    where .
    As a result, we get formulas
    tangential acceleration: ;
    normal acceleration: .

Dynamics

Dynamics is a branch of theoretical mechanics that deals with mechanical movement material bodies, depending on the causes that cause them.

    Basic concepts of dynamics
  • inertia- this is the property of material bodies to maintain a state of rest or uniform rectilinear motion until external forces change this state.
  • Weight is a quantitative measure of the inertia of a body. The unit of mass is kilogram (kg).
  • Material point is a body with a mass, the dimensions of which are neglected in solving this problem.
  • Center of mass of a mechanical system is a geometric point whose coordinates are determined by the formulas:

    where m k , x k , y k , z k- mass and coordinates k- that point of the mechanical system, m is the mass of the system.
    In a uniform field of gravity, the position of the center of mass coincides with the position of the center of gravity.
  • Moment of inertia of a material body about the axis is a quantitative measure of inertia during rotational motion.
    The moment of inertia of a material point about the axis is equal to the product of the mass of the point and the square of the distance of the point from the axis:
    .
    The moment of inertia of the system (body) about the axis is equal to the arithmetic sum of the moments of inertia of all points:
  • The force of inertia of a material point is a vector quantity equal in absolute value to the product of the mass of a point and the module of acceleration and directed opposite to the acceleration vector:
  • Force of inertia of a material body is a vector quantity equal in absolute value to the product of the body mass and the module of acceleration of the center of mass of the body and directed opposite to the acceleration vector of the center of mass: ,
    where is the acceleration of the center of mass of the body.
  • Elemental Force Impulse is a vector quantity equal to the product of the force vector by an infinitesimal time interval dt:
    .
    The total impulse of force for Δt is equal to the integral of elementary impulses:
    .
  • Elementary work of force is a scalar dA, equal to the scalar

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 those concepts that were introduced in section Kinematics, here it is necessary to use new concepts that reflect the specifics of the impact of forces on various bodies and the response of bodies to these impacts. Let's consider the main of these concepts.

a) strength

Force is the quantitative result of the impact on a given body by 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 line MN on which the force vector is located is called line of force. The length of the force vector, measured on a certain scale, is called numerical value or modulus of the force vector. The modulus of force is denoted as or . The action of a force on a body is manifested either in its deformation, if the body is stationary, or in imparting acceleration to it when the body moves. On these manifestations of force, the device of various instruments (force meters or dynamometers) for measuring forces is based.

b) system of forces

The considered set of forces forms force system. 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 influence 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 has the same effect on a free body as some system of forces, then this force is called resultant of this system of forces. This is written as follows:

,

which means equivalence the impact on the same free body of the resultant and some system of n forces.

Let us now turn to the consideration of more complex concepts related to the quantitative determination of the rotational effects of forces.

e) moment of force relative to a point (center)

If the body under the action of a 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 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 factor vectors (in this case, the plane of the triangle OAB), directed in the direction from which the shortest turn of the first factor vector to the second factor vector visible against the clock (Fig. 2). With this order of the vectors of the factors of the cross product (1), the rotation of the body under the action of the force will be visible against the clock (Fig. 2). Since the vector is perpendicular to the plane of the force, its location in space determines the position of the plane of the force. The numerical value of the vector of the moment of force relative to the center is equal to twice the 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 arm of the force.

If the position of the plane of action of the force in space is not essential for characterizing 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, 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 and its shoulder, taken with a plus or minus sign. In this case, a positive moment corresponds to the rotation of the body under the action of a given force against the clock, and a negative moment corresponds to the rotation of the body in the direction of the clock. From formulas (1), (2) and (3) it follows that the moment of force relative to a point is equal to zero only if the arm of this forcehzero. Such a force cannot rotate the body around a given point.

f) Moment of force about the axis

If a body under the action of a 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 a physical quantity is introduced to quantify this rotational effect, which is called moment of force about a given axis.

z

b Fxy

Figure 3 shows a diagram in accordance with which the moment of force about 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 according to the stereometry theorem on the projection of a flat figure onto a given plane, we have:

where the plus sign corresponds to a positive value of cos, i.e. sharp corners, and the minus sign corresponds to the negative value of cos, i.e. obtuse angles , due to the direction of the vector . In turn, SO ab=1/2abh, where h ab . The value of the segment ab is equal to the force projection onto the xy plane, i.e. . ab = F xy .

Based on the foregoing, as well as equalities (4) and (5), we determine the moment of force about the z-axis as follows:

Equality (6) allows us to formulate the following definition of the moment of force about any axis: The moment of force about a given axis is equal to the projection onto this axis of the vector of the moment of this force relative to any point of a given axis and is defined as the product of the force projection onto a plane perpendicular to the given 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. Otherwise, the moment of force about the axis is taken as negative. Since this definition of the moment of force relative to the axis is quite difficult to remember, it is recommended to remember the formula (6) and Fig. 3, which explains this formula.

From formula (6) it follows that 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 equal to zero), or the line of action of the force intersects the axis (then the projection arm h=0). This fully corresponds to the physical meaning of the moment of force about the axis as a quantitative characteristic of the rotational action of force on a body with an axis of rotation.

g) body weight

It has long been noted that under the influence of a force, the body picks up speed gradually and continues to move if the force is removed. This property of bodies, to resist a change in their motion, was called inertia or inertia of bodies. The quantitative measure of the inertia of a body is its mass. Besides, body mass is a quantitative measure of the effect of gravitational forces on a given bodythe greater the mass of the body, the greater the gravitational force acts on the body. As will be shown below, uh These two definitions of body weight are related.

Other concepts and definitions of dynamics will be discussed later in the sections where they first occur.

2. Bonds and reactions of bonds

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 located on the table surface can move in any direction, except for the direction perpendicular to the table surface down. Hinged doors can rotate, but cannot move forward, etc. Bodies that cannot move in space in one direction or another are called not free.

Everything that limits the movement of a given body in space is called bonds. These can be some other bodies that prevent the movement of this body in some directions ( physical connections); more broadly, it may be some conditions imposed on the movement of the body, limiting this movement. So, you can set a condition for the movement of a material point to occur along a given curve. In this case, the connection is specified mathematically in the form of an equation ( connection equation). The question of the types of links will be considered in more detail below.

Most of the bonds imposed on bodies are practically physical bonds. Therefore, the question arises about the interaction of a given 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, during the interaction of a given body and a connection, 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 the given body. This last power is called bond reaction force or simply, connection reaction.

When solving practical problems of dynamics, it is necessary to be able to find the direction of reactions various types connections. The general rule for determining the direction of a bond reaction can sometimes help with this: The reaction of a bond is always directed opposite to the direction in which this bond prevents the movement of a given body. If this direction can be specified definitely, then the reaction of the connection will be determined by the direction. Otherwise, the direction of the bond reaction is indefinite and can only be found from the corresponding equations of motion or equilibrium of the body. In more detail, the question of the types of bonds and the direction of their reactions should be studied according to the textbook: S.M. Targ A short course in theoretical mechanics "Higher school", M., 1986. Ch.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 are, in fact, 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 connections of lectures on philosophy at home. Lectures were... social psychology and ethnopsychology. 3. Theoretical results In social Darwinism were ...

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