Practical work on the topic: "Work and power during rotational movement"

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Brief theory:

The work of a constant force applied to a rotating body

Imagine a disk rotating around a fixed axis under the action of a constant forceF (fig. 6) , the application point of which moves with the disc. Expand the forceF into three mutually perpendicular components:F 1 - circumferential force,F 2 - axial force,F 3 - radial force.

When turning the disk at an infinitely small angledφ forceF will perform elementary work, which, based on the theorem on the work of the resultant, will be equal to the sum of the work of the components.

It is obvious that the work of the componentsF 2 andF 3 will be equal to zero, since the vectors of these forces are perpendicular to the infinitesimal displacementds application pointsM , so the elementary work of forceF equal to the work of its componentF 1 :

dW \u003d F 1 ds \u003d F 1 Rdφ .

When turning the disc to the end angleφ work forceF equals

W \u003d ∫ F 1 Rdφ \u003d F 1 R ∫ dφ \u003d F 1 Rφ ,

where is the angleφ expressed in radians.

Since the moments of the constituentsF 2 andF 3 about the axisz are equal to zero, then based on moment of powerF about the axisz is equal to:

M z (F) \u003d F 1 R .

The moment of force applied to the disc relative to the axis of rotation is called the torque, and according to the standardISO , denoted by the letterT :

T \u003d M z (F) , Consequently,W \u003d Tφ .

The work of a constant force applied to a rotating body is equal to the product of torque and angular displacement .

An example of solving the problem

Task: the worker rotates the winch handle by forceF \u003d 200 N perpendicular to the radius of rotation.
Find work spent over timet \u003d 25 seconds if the length of the handler \u003d 0.4 m , and its angular velocityω \u003d π / 3 rad / s .

Decision.
First of all, we define the angular displacementφ winch handles25 seconds :

φ \u003d ωt \u003d (π / 3) × 25 \u003d 26.18 rad.

W \u003d Tφ \u003d Frφ \u003d 200 × 0.4 × 26.18 ≈ 2100 J ≈ 2.1 kJ .

The power of a force applied to a uniformly rotating body is equal to the product of torque and angular velocity .

If the work is done by a force applied to a uniformly rotating body, then the power in this case can be determined by the formula:

P \u003d W / t \u003d Tφ / t orP \u003d Tω .

Option number 1

On two cords of equal length, equal to 0.8 m, two lead balls weighing 0.5 and 1 kg are suspended. The balls are in contact with each other. The ball of smaller mass was taken to the side so that the cord deflected by an angle α \u003d 60 °, and released. How high will both balls rise after collision? The blow is considered central and inelastic. Determine the energy expended on the deformation of the balls upon impact.

The flywheel weighing 4 kg rotates freely around the horizontal axis passing through its center at a frequency of 720 min-1. The mass of the flywheel can be considered distributed over its rim with a radius of 40 cm. After 30 s, the flywheel stopped under the influence of the braking torque. Find the braking torque and the number of revolutions that the flywheel makes to a complete stop.

A body with a mass of m \u003d 1.0 kg falls from a height of h \u003d 20 m. Neglecting the air resistance, find the average power developed by gravity along the path h and the instantaneous power at a height of h / 2.

Option number 2

The flywheel rotates according to the law expressed by the equation, where A \u003d 2 rad, B \u003d 32 rad / s, C \u003d -4 rad / s2. Find the average powerNdeveloped by the forces acting on the flywheel during its rotation, until it stops, if the moment of inertia I \u003d 100 kg m 2 .

A body of mass m rotates on a horizontal surface along a circle of radius r \u003d 100mm. Find the work of the friction force when the body is rotated through the angle α \u003d 30. The coefficient of friction between the body and the surface is k \u003d 0.2.

The first ball with a mass of m1 \u003d 2 kg is moving at a speed whose value is v1 \u003d 3 m / s. The second ball with a mass of m2 \u003d 8 kg is moving at a speed of v2 \u003d 1 m / s. Find speedv 1 first ball and speedv 2 the second ball immediately after the impact, if: a) the balls move towards each other; b) the first ball overtakes the second. Consider the impact as central and absolutely elastic.

Consider the formulas for determining the work and power of the force applied at any point in a rigid body performing translational or rotational motion.

1. Work and power of the force applied to a rigid body in translational motion.

Consider a rigid body moving in translation with respect to the inertial reference frame under the action of a force applied at an arbitrary point (Fig. 24).

In the case of the translational motion of a rigid body, all its points move with speeds the same in magnitude and direction. Let's designate the speed of the body.

Using formula (4.31), we obtain

where is the differential of the radius vector of an arbitrary point of the rigid body.

Figure: 24. Translational motion of a rigid body under the action of force

Dividing (4.49) by dt, we obtain an expression for determining the power of the force acting on the body in translational motion:

where is the angle between the vectors of the velocity force.

That is, the power of the force during the translational motion of a rigid body is defined as the scalar product of the force vector by the velocity vector of the rigid body.

Integrating (4.49) on some finite displacement of the point M from starting position M 0 to position M 1, we get the full work of the force acting on the body on this displacement

2. The work and power of the force applied to a rigid body that rotates.

Consider the rotation of a rigid body around a fixed vertical axis Oz under the action of a force applied at an arbitrary point of this body M (fig. 25).

Figure: 25. Rotation of a rigid body around a fixed axis

Point position M in axes Oxyz defined by the radius vector. Point speed M directed tangentially to the trajectory of motion (a circle centered on the axis of rotation). The vector of this velocity can be determined by the vector Euler formula, known from the course of rigid body kinematics

where is the vector of the angular velocity of rotation of the rigid body.

Using formula (4.32), we obtain

Changing the factors in the mixed vector product in a circular order, we get

where is the vector moment of force, relative to the center O.

The angle between the vectors of moment and angular velocity.

Considering that:

1. - moment of force, relative to the axis of rotation Oz.

2.and therefore,

finally get

In this way, the elementary work of a force applied at any point of a rigid body rotating around a fixed axis is equal to the product of the moment of this force relative to the axis of rotation by the differential of the angle of rotation of the body.

To determine the total work of the force when the body rotates through an angle φ, integrating expression (4.53), we obtain

In the case when, the complete work can be determined by the formula

where φ is the angle of rotation of the body at which the work of the force is determined.

If the direction of the moment and the angular velocity coincide, then the work of the force is considered positive, otherwise - negative.

Let us determine the power of the force when a rigid body rotates around an axis. Using formula (4.40), we obtain

I.e the power of the force applied to a rotating rigid body is defined as the product of the moment of force about the axis of rotation by the angular velocity of the body ... The power sign is determined similarly to the work sign.

nality (∂ f ∂ ϕ) 2. From this it is seen that the coefficient of inertia of the object depends

sieve from the choice of the generalized coordinate and can be recalculated.

The FE of a nonstationary holonomic one-degree system has the structure

round of a square polynomial with respect to the generalized velocity q &, the coefficient

whose numbers generally depend on q and t:

2T \u003d aq & 2 + 2a 1 q & + 2a 0, for a \u003d a (q, t), a 1 \u003d a 1 (q, t), a 0 \u003d a 0 (q, t) (5.10)

The dimension of the coefficients a, a 0, a 1 is determined by L. Euler's principle: all terms in the expressions must have the same dimension.

5.3. Power of force

The area of \u200b\u200bspace in which a force is applied to a material object is called vector force field... This area can be three-dimensional (for example, spherical), or two-dimensional, or represent a segment of a straight or curved line. It is usually considered that the force depends only on the coordinates (x, y, z) of the point of application of the force, or on one or two coordinates, or is constant in magnitude and direction. Cases are also allowed when the forces depend both on the speed of a point and on time, i.e. the force is given in the area of \u200b\u200bspace of coordinates, velocities, time. There are cases when

where the force depends on the acceleration.

at an instant t in the Oxyz frame is called

Power of force F

scalar equal to the dot product of the force

on the speed of the point

force v in this system:

m / s \u003d W)

Fv cos (F, v)

Zz, (H

According to this definition, the power of a force is a positive scalar if the angle between the force and the speed is sharp (in this case, the force promotes movement, an increase in kinetic energy) and negative if the angle is obtuse (when the force slows down the movement). The power of the force is zero if the force is perpendicular to the speed of the point of application of the force, or if the point of application of the force has no speed.

The powers in the two frames of reference are different if the systems move relative to one another, therefore, the frame of reference in which the power of the forces is calculated should be indicated.

The power of the friction forces, as well as of other dissipative forces directed against the motion, is negative.

The traction power of the wheel (if there is no wheel slip) is zero, since the point of application of the force has no speed.

Consider the case when the forces depend only on the position of the point at

U (x, y, z) is a function of the position of the point of application of the force, i.e. - a function of Cartesian (or generalized) coordinates. In this case, the force F (x, y, z) is called potential, and the “force function” U with the opposite sign is called

potential energy: П (x, y, z) \u003d - U (x, y, z). The area of \u200b\u200bspace in which

second, a potential force acts on the body, called potential force field... Any constant can be added under the sign of the derivative; therefore, the strength function and potential energy are determined accurate to a constant that determines the reference level. In the general case, the potential energy can be defined as a function П (q 1, ..., q n), obtained

by transforming the power to the form: P \u003d - П & (q 1, ..., q n), where q s - generalized

coordinates.

Let the body move arbitrarily in space, i.e. it moves with the pole O at a speed v O and rotates at an angular speed ω.

The power of a pair of forces applied to a solid does not depend on the speed of the pole. It is equal to the scalar product of the moment of the pair of forces and the angular velocity.

P \u003d M			M ω cos (M, ω			) \u003d M xω x + M yω y + M zω z,

where M is the moment of a pair of forces, ω is the angular velocity of a rigid body, which, as is known, does not depend on the choice of the pole. The power of dissipative pairs of forces is negative. The power of a pair of forces does not depend on the place of its application to the body. The power of a pair of frictional forces in the bearing is negative, since the frictional moment and the angular velocity of rotation are oppositely directed.

The power of the system of forces applied to a rigid body is equal to the scalar product of the principal vector R of the system and the velocity of any pole of the body, added with the scalar product of the principal moment M 0 of forces relative to this pole by the angular velocity of the body:

v O + M

O ω

for R \u003d ∑ F i, M O \u003d ∑ r i × F i.

5.4. Work and potential energy

The elementary work of a force in the selected coordinate system Oxyz (stationary or moving) is called an infinitesimal quantity equal to the scalar product of the force and the elementary movement of the point of application of the force in this system:

d ′ A \u003d F		d r \u003d Xdx + Ydy + Zdz \u003d F | d r | cos (F, d r), (N m \u003d J)

Here, d ΄A denotes the infinitely small work done by the force in an infinitely small time interval, d r is an elementary movement co-directed with the speed of a point. The dash marks that d ΄A is not always the total differential of some function.

Obviously, the product Pdt is equal to the elementary work d ΄A:

The power multiplied by a small time interval ∆t is an approximate value of the work ∆A of the force during this interval, the power is approximately equal to the work of the force for 1 sec. The work of force for a finite time interval is a definite integral of power over time:

A12 \u003d ∫ Pdt \u003d ∫			v dt for v \u003d r & \u003d dr / dt.

To calculate the work according to this general formula, it is necessary to know the power as a function of time or the force and speed as functions of time t only. But in some special cases (the case of a potential force, the case of a constant friction force with a constant direction of motion), it is possible to calculate the work without using the kinematic equations of motion of the point of application of the force, it is enough to know only the initial and final position of the point.

Consider the motion of the point of application of force in relation to two frames of reference moving relative to one another. The speed of the point in the two systems is different, therefore the power of the force will be different. Thus, the concepts of power, work, are formulated in relation to a specific frame of reference, mainly in relation to IFR or PSO (inertial or translational frames of reference).

Definition Force F is called potential, and its force field -

potential force fieldif two conditions are met:

1) The force satisfies one of the following conditions: the force is constant in magnitude and direction F \u003d const or depends only on the coordinates of the point (all three or part) of its application, i.e. F \u003d F (x, y, z).

2) The elementary work d ′ A of the force is the total differential of some function of coordinates, or the power of the force at any time is equal to the total time derivative of some function Π (x, y, z)

The function P (x, y, z), obtained by transforming the expression of elementary work, or from the expression of power, is called

potential energy of the potential force field at the point M (x, y, z).

Thus, the vector force field of the force F (x, y, z) is associated with

a mathematically simpler field of a scalar function of three variables P (x, y, z), either - a function of two variables P (x, y), or - a function of one variable P (x)

Potential energy can be represented not only in a Cartesian coordinate system, but also in a cylindrical, spherical coordinate system, in general, it is a function of some generalized coordinates

nat P (q 1, q 2, q 3).

The surfaces defined by the equation П (q 1, q 2, q 3) \u003d C, where C is an arbitrarily assigned constant parameter, are called equipotential surfaces.

Note that under the sign of the differential it is always possible to add or subtract any constant, so that the function in formula (5.18) is determined up to a constant. The constant is arbitrarily assigned, for example, set equal to zero, thereby choosing the reference level of the family of equipotential surfaces.

The power of the potential force is equal to that taken with a minus sign

water in time from potential energy P \u003d −Π &. Let us substitute this expression in the definite integral (5.17). Let us obtain an expression for the work of the potential force on the final displacement of the point of application of the force, carried out in a finite period of time:

A 12 \u003d P (x 1, y 1, z 1) - P (x 2, y 2, z 2) \u003d P1 - P2.

Thus, the work of a potential force when it moves for an

the interval from point M 1 (x 1, y 1, z 1) to point M 2 (x 2, y 2, z 2) along any trajectory is equal to the decrease in potential energy on this displacement, i.e. equal to different

of the potential energies at the first and second points of the potential field. The work of the potential force does not depend on the shape of the trajectory connecting two points. In particular, the work of the potential force on any closed trajectory is equal to zero, and the work when the point of application of the force passes from the equipotential surface P \u003d C1 to the surface P \u003d C2 is equal to

sti constants: A12 \u003d C1 -C2.

Special case As an initial point M 1 (x 1, y 1, z 1) we take any point M (x, y, z) of the potential field, and as M 2 (x 2, y 2, z 2) we take such a point field M (x O, y O, z O), in which the potential energy is taken equal to

We get the following physical interpretation. The potential energy at any point M of the potential field is equal to the work of the applied force when its point of application moves from position M along any smooth or nonsmooth trajectory to a position in which the potential energy is taken to be zero, and is also equal to the work of the force taken with a minus sign on displacement in the position M (x, y, z) from the “zero” position, in which the potential energy is taken to be zero.

Example 1 Let us find the potential energy of gravity G \u003d - Gk,

oxyz system Oxyz directed with the unit vector k of the vertical axis Oz. By the method of elementary work, we get:

d ΄A \u003d G x dx + G y dy + G z dz \u003d –Gdz \u003d - d (Gz) \u003d\u003e П \u003d Gz.

Using the power method, we obtain

P \u003d G x x & + G y y & + G z z & \u003d −Gz & \u003d - (Gz) Π \u003d Gz.

Thus, the potential energy of gravity is equal to the product of the weight of a material point by the height of the point M above the Oxy plane, satisfying the condition z \u003d 0. Here, the Oxy plane is assigned

zero equopotential plane. The potential energy of gravity is negative at points located under the plane Oxy, at z< 0. На любых горизонтальных плоскостях данная потенциальная энергия одинакова во всех точках, т.е. горизонтальные плоскости являются эквипотенциальными поверхностями. Работа силы тяжести на перемещении с плоскости уровня z = z 1 на плоскость z = z 2 определяется по формуле:

A 12 \u003d P1 - P2 \u003d G (z 1 - z 2) \u003d ± Gh for h \u003d | z 1 –z 2 |.

This work is proportional to the difference (decrease) of levels, it is negative if the first level is lower than the second.

Remark. If the Oz axis is directed downward, we obtain a formula with the opposite sign: П \u003d –Gz.

Example 2. Potential energy of the spring force. The force field of the horizontal spring has the form of the horizontal axis Ox. The beginning of the axis is compatible with the free end of an undeformed spring, x is the deformation of the tension of the spring at x\u003e 0, or the compression of the spring at x< 0. Упругая сила пружины F = − cxi , где i - орт оси x . Она всегда направлена противоположно деформации. Методом мощности находим потенциальную энергию силы упругости

P \u003d Fx x \u003d - c x x \u003d - (c x					Π \u003d cx

Let's imagine that a spring is very slowly stretched by an external force,

slowly increasing from zero to the value F nn \u003d cxi. We assume that at each moment of time the elastic force of the spring balances the external force.

The average value of the magnitude of the force F vn in the interval is: F cp \u003d cx / 2.

The elastic force of the spring, while doing negative work to resist stretching, stores in the spring a positive potential

energy equal to Π \u003d F x \u003d cx 2/2.
Elastic force work on deformation				X 2 - x 1 equals A 12 \u003d (x 2 2 - x 1 2) c / 2.
Obviously, A 12< 0 при x1 < x2 и A 12 > 0 for x1\u003e x2
	3. Earth's gravity			according to the inverse square law:
F \u003d γ m m / r2,			\u003d - γ m m r / r 3, where r is the radius vector of the material point at

geocentric frame of reference, γ \u003d 6.672 · 10-11 (m3 / (kg · s2) - constant gravity

gotenia, r / r \u003d e is the unit vector of the radius vector of the body (material point) drawn from the center of the Earth, m 1 \u003d 61024 (kg) is the mass of the Earth, m is the mass of the body, γm 1 \u003d

3986 · 1011 (m3 / s2) - geocentric gravitational constant. Considering

identities r r \u003d r 2,

γ m1 m

γ m1 m

γ m1 m

γ m1 m

d A \u003d -

r dr \u003d -

dr \u003d d (-

Π (r) \u003d -

Note that П (r) → 0 as r → ∞, therefore, the potential energy

at infinity is taken to be zero.

"

In the section "Kinematics" it is established that the speed of any point of a rigid body is geometrically summed up from the speed of the point taken as a pole and the speed obtained by the point during the spherical motion of the body around the pole. In dynamics, the center of mass of the body is always taken as the pole. The speed of any point of the body is determined by the formula

- the speed of the center of mass of the body;

- the vector of the instantaneous angular velocity of the body;

- radius vector in relation to the center of mass of the body.

For the power of the force applied to an absolutely rigid body, we get:

Of particular interest is the plane-parallel motion of a rigid body. In this important special case, the power of the force can be calculated by the formula:

where is the angle between the force and velocity vectors of the body's center of mass.

End of work -

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All topics in this section:

Basic laws of mechanics
Theoretical mechanics is one of the so-called axiomatic sciences. It is based on a system of starting positions - axioms, accepted without proof, but verified not only by direct

Axiom 3
Two material points interact with forces equal in magnitude and directed along one straight line in opposite directions (Fig.!. 2). Axiom 4 (Principle

Point speed
The speed of movement of a point is characterized by its speed, to the definition of which we now turn. Let the moment in time

Point acceleration
The rate of change of the velocity vector is characterized by the point acceleration. Let at the moment of time point nah

Axiom 3
A system of two forces applied to an absolutely rigid body is balanced (equivalent to zero) if and only if these forces are equal in magnitude and act along one straight line in opposite

Moment of force relative to a point
Let the force be given at the point

Moment of force about the axis
The moment of force relative to the axis is the projection onto the axis of the moment of force calculated relative to any point of this axis:

A couple of forces
A pair of forces is a system of two forces equal in magnitude and acting along parallel straight lines in opposite directions. The plane in which

Differential equations of motion of a mechanical system
Consider a mechanical system consisting of material points. For each point of the system in the inertial system about

Basic properties of internal forces
Consider any two points of a mechanical system and

The theorem on the change in the momentum of a mechanical system
Let us sum up all equalities (3.1) term by term: Taking into account the first basic relation

The theorem on the change in the angular momentum
Let us multiply each of the equations (3.1) on the left vector by the radius-vector of the corresponding point and add

Equilibrium conditions
Let us dwell on the issues of equilibrium of material bodies, which constitute an essential part of the section "Statics" of the course of theoretical mechanics. Balance in mechanics is traditionally

Equilibrium of a system of forces whose lines of action lie in the same plane
In many cases of practical interest, the body is in equilibrium under the action of a system of forces, the lines of action of which are located in the same plane. We take this plane as the coordinate

Farms calculation
Calculation of trusses occupies a special place among static tasks. A truss is a rigid structure of rectilinear bars (Figure 3.3). If all the rods of the truss and all attached to it

Body balance in the presence of friction
As you know, when a body slides on a support surface, resistance arises that inhibits sliding. This phenomenon is taken into account by introducing the friction force into consideration.

Center of Parallel Forces
This concept is introduced for a system of parallel forces having a resultant, and the points of application of the forces of the system are points

Body center of gravity
Consider a material body located near the surface of the Earth (in the field of gravity). Let us first assume that the body consists of a finite number of material points, in other words, particles,

Center of gravity of a mechanical system. The theorem on the motion of the center of mass
The inertial properties of a material body are determined not only by its mass, but also by the nature of the distribution of this mass in the body. An important role in describing such a distribution is played by the position of the center

LECTURE 5
5.1. Motion of an absolutely rigid body One of the most important tasks of mechanics is the description of the motion of an absolutely rigid body. In general, different points

The translational motion of a rigid body
The motion of a rigid body is called translational, in which any straight line drawn in the body remains parallel to its original position during the entire movement.

Kinematics of the rotational motion of a rigid body
During rotational motion in the body, there is a single straight line, all points of which

Body speed
Finally, we obtain: (5.4) Formula (5.4) is called Euler's formula. Figure 5.

The differential equation of the rotational motion of a rigid body
The rotation of a rigid body, like any other movement, occurs as a result of the action of external forces. To describe the rotational motion, we use the theorem on the change in the angular momentum with respect to

Kinematics of plane-parallel motion of a rigid body
The movement of a body is called plane-parallel if the distance from any point of the body to some fixed (main) plane remains unchanged throughout the movement

Differential equations for plane-parallel motion of a rigid body
When studying the kinematics of the plane-parallel motion of a rigid body, any point of the body can be taken as a pole. When solving problems of dynamics, the center of mass of the body is always taken as a pole, and

Koenig system. Koenig's first theorem
(Study on your own) Let the frame of reference be fixed (inertial). System

Work and power of force. Potential energy
Half of the product of the mass of a point by the square of its velocity is called the kinetic energy of the material point. The kinetic energy of the mechanical system is called

The theorem on the change in the kinetic energy of a mechanical system
The theorem on the change in kinetic energy is one of the general theorems of dynamics, along with the previously proven theorems on the change in the momentum and the change in the moment of quantities

The work of internal forces of a geometrically unchanging mechanical system
Note that, in contrast to the theorem on the change in the momentum and the theorem on the change in the angular momentum, the theorem on the change in the kinetic energy generally includes internal forces.

Calculation of the kinetic energy of an absolutely rigid body
Let us obtain formulas for calculating the kinetic energy of an absolutely rigid body for some of its movements. 1. With translational motion at any time, the speed of all points of the body is one

Work of gravity
When calculating the work of gravity, we will assume that we are considering a limited region of space near the surface of the Earth, the dimensions of which are small compared to the dimensions of the Earth

Elastic force work
Elastic force is usually associated with the response of a linear elastic spring. Let us direct the axis along pr

Torque work
Let the force be applied at some point of the body with the axis of rotation. The body rotates at an angular rate

Possible speeds and possible movements
Let us first introduce the concepts of possible velocity and possible displacement for a material point on which a holonomic holding nonstationary constraint is imposed. Possible checkmate speed

Perfect connections
The constraints imposed on a mechanical system are called ideal if the sum of the work of all constraint reactions on any possible displacement of the system is zero:

The principle of possible movements
The principle of possible displacements establishes the equilibrium conditions for mechanical systems. The equilibrium of a mechanical system is traditionally understood as the state of its rest in relation to the selected inertial

General equation of dynamics
Consider a mechanical system consisting of material points, on which are superimposed ideal ude

The work of internal forces on the final displacement is zero.

The work of a force acting on a translationally moving body is equal to the product of this force and the increment of linear displacement.

The work of the force acting on the rotating body is equal to the product of the moment of this force relative to the axis of rotation by the increment in the angle of rotation:; ... Power:
.

Kinetic energy of a mechanical system for various types of motion.

Kinetic energy of a mechanical system - scalar equal to the sum of kinetic energies of all points of the system: .

When moving forward:

Rotational motion:

For plane-parallel motion:, where d is the distance from the center of mass to the MCS

27. Theorem about the change in the kinetic energy of a material point.

Kinetic energy of a material point is a scalar equal to half the product of the mass of a point by the square of its velocity.

The basic equation of dynamics: , multiply by the elementary displacement: ; ; ... Integrating the resulting expression:

Theorem: the change in the kinetic energy of a material point at a certain displacement is equal to the work of the force acting on the point at the same displacement.

Theorem about the change in the kinetic energy of a mechanical system.

Since the work of internal forces is zero, then:
.

Theorem: the change in the kinetic energy of a mechanical system at a final displacement is equal to the sum of the work of external forces at the same displacement.

The principle of possible movements for a mechanical system.

; , let the bonds imposed on the points of the mechanical system are two-sided, stationary, holonomic and ideal, then:.

The principle of possible displacements - lagrange principle - for equilibrium of a mechanical system with two-sided, stationary, holonomic and ideal constraints, it is necessary and sufficient that the algebraic sum of the work of the given forces on a possible displacement equals zero.

D'Alembert's principle for a material point.

The geometric sum of all forces applied to a moving material point and forces of inertia of this point is equal to zero

The d'Alembert principle for a non-free mechanical system.

In a moving non-free mechanical system for each material point at any moment of time, the geometric sum of the assigned forces applied to it, coupling reactions and inertial forces is equal to zero. Multiplying both parts of the expression by r i we get:;
.

, the sum of the moments of the given forces, the coupling reactions and the inertial forces about the coordinate axes is equal to zero.

Reducing the forces of inertia of points of a rigid body to the simplest form.

To the system of inertia forces of points of a rigid body, you can apply the Punch method, considered in statics. Then any system of inertial forces can be reduced to the main vector of inertial forces and the main moment of inertia forces.

In translational motion: Ф \u003d -ma (with translational motion of a rigid body, the forces of inertia of its points are reduced to the main vector of inertia forces equal in magnitude to the product of the body's mass, to the acceleration of the center of mass applied at this center and directed towards the opposite acceleration of the center of mass).

During rotational motion: М \u003d -Iε (during rotational motion of a rigid body, the forces of inertia of its points are reduced to the main moment of inertia forces equal to the product of the moment of inertia of the body relative to the forces of rotation by angular acceleration. This moment is directed towards the opposite angular acceleration).

For plane motion: Ф \u003d -ma M \u003d -Iε (for plane motion of a rigid body, the inertial forces of its points are reduced to the main vector and the main moment of inertia forces).

General equation of dynamics. D'Alembert-Lagrange principle.

D'Alembert principle: å (P i + R i + Ф i) \u003d 0; е (P i + R i + Ф i) Dr i \u003d 0, we put. that the constraints imposed on the mechanical system are two-sided, stationary, holonomic and ideal, then: å (R i × Dr i) \u003d 0;

е (P i + Ф i) Dr i \u003d 0 - general equation of dynamics - for the movement of a mechanical system with two-sided, stationary, holonomic and ideal constraints, the sum of the work of the given forces and the forces of inertia of the points of the system at any possible displacement is equal to zero.