nality (∂ f ∂ ϕ ) 2 . This shows that the coefficient of inertia of the object depends

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

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

round of the quadratic polynomial with respect to the generalized speed q & , coefficient

the values ​​of which generally depend on q and t:

2T = aq & 2 + 2a 1 q & + 2a 0 , with a = a (q ,t ), a 1 = a 1 (q ,t ), a 0 = a 0 (q ,t ) (5.10)

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

5.3. Power power

The region of space 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 believed that 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 on both the speed of the point and time, i.e. the force is specified in the area of ​​space of coordinates, velocities, and time. There are cases where

where force depends on acceleration.

at instant t in the reference frame Oxyz is called

Power power F

scalar equal to the dot product of the force

applied to the speed of the point

force v in this system:

m/s=W)

Fv cos(F ,v )

Zz, (N

According to this definition, the power of a force is a positive scalar if the angle between force and speed is acute (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 reference systems are different if the systems move relative to each other, so the reference system in which the power of the forces is calculated should be indicated.

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

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

Let us consider the case when the forces depend only on the position of the point of

U (x, y, z) is a function of the position of the force application point, i.e. – 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: P (x, y, z) = − U (x, y, z). The region of space in which

which a potential force acts on a body is called potential force field. Under the derivative sign, you can add any constant, so the force function and potential energy are determined up to a constant that determines the reference level. In general, potential energy can be defined as a function P (q 1,..., q n) obtained

by transforming the power to the form: P = − П & (q 1 ,..., q n ) , where q s is a generalized

new coordinates.

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

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

P = M			M ω cos(M ,ω			) = 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 pole. The power of dissipative force pairs is negative. The power of a pair of forces does not depend on the place where it is applied to the body. The power of a pair of friction forces in the bearing is negative, since the friction torque and the angular velocity of rotation are opposite directions.

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

vO+M

Oω

for R = ∑ F i , M O = ∑ r i × F i .

5.4. Work and potential energy

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

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

Here d ΄A denotes the infinitesimal work done by a force in an infinitesimal time interval, d r is the elementary displacement co-directed with the speed of the point. The prime indicates that d ΄A is not always a complete differential of some function.

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

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 in 1 second. The work done by a force over a finite time interval is called a definite integral of power over time:

A12 = ∫ Pdt = ∫			v dt for v = r & = dr / dt .

To calculate work using this general formula, it is necessary to know power as a function of time or force and speed as functions of only time t. But in some special cases (the case of potential force, the case of constant friction force with a constant direction of movement), 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.

Let us consider the movement of the point of application of the force in relation to two reference systems moving one relative to the other. 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 and work are formulated in relation to a specific reference system, mainly in relation to ISO or PSO (inertial or translational reference systems).

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

potential force field, if two conditions are met:

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

2) The elementary work d ′ A of a 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 point M(x, y, z).

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

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 the Cartesian coordinate system, but also in cylindrical, spherical coordinate systems; in general, it is a function of some generalized coordinates.

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

Surfaces defined by the equation P(q 1, q 2, q 3) = C, where C is an arbitrarily assigned constant parameter, are called equipotential surfaces.

Note that under the differential sign you can always add or subtract any constant, so that the function Π 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 the product taken with a minus sign

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

A 12 = P(x 1, y 1, z 1) – P(x 2, y 2, z 2) = P1 – P2.

Thus, the work of a potential force when it moves behind an in-

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 loss of potential energy during this movement, i.e. equal to different

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

sti constants: A12 = C1 - C2.

Special case As the 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 obtain 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 moving its point of application from position M along any smooth or non-smooth trajectory to a position in which the potential energy is taken equal to zero, and is also equal to the work of the force taken with a minus sign on the displacement in position M (x,y,z) from the “zero” position, in which the potential energy is taken equal to zero.

Example 1 Let us find the potential energy of gravity G = − Gk, pro-

oppositely directed with the unit vector k of the vertical axis Oz of the Oxyz system. Using the elementary method we obtain:

d ΄A = G x dx + G y dy + G z dz = –Gdz = – d (Gz) => П = Gz.

Using the power method we obtain

P = G x x & +G y y & +G z z & = −Gz & = −(Gz ) Π = Gz .

Thus, the potential energy of gravity is equal to the product of the weight of the material point and the height of the location of the point M above the Oxy plane, satisfying the condition z = 0. Here the Oxy plane is assigned

zero equipotential plane. Gravity potential energy is negative at points located below the Oxy plane, 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 = P1 – P2 = G (z 1 – z 2 ) = ± Gh at h = |z 1 –z 2 |.

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

Note. If the Oz axis is directed downward, we obtain a formula with the opposite sign: P = –Gz.

Example 2. Potential energy of the elastic force of a spring. The force field of a horizontal spring has the form of a horizontal axis Ox. The origin of the axis is compatible with the free end of the undeformed spring, x is the tensile strain of the spring at x > 0, or the compression strain 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 = Fx x = − c x x = − (c x					Π = cx

Let us imagine that the spring is stretched very slowly by an external force,

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

The average value of the force F ext over the interval is equal to: F cр = cx / 2.

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

energy equal to Π = F x = cx 2 / 2.
Work of elastic force on deformation				X 2 − x 1 is equal to A 12 = (x 2 2 – x 1 2 )c /2.
Obviously A 12< 0 при x1 < x2 и A 12 >0 for x1 > x2
	3. Earth's gravity			according to the inverse square law:
F = γ m m / r2 ,			= − γ m m r / r 3 , where r is the radius vector of the material point in

geocentric reference system, γ = 6.672 10–11 (m3 /(kg s2) - constant gravity

goteny, r / r = e - ort of the radius vector of the body (material point) drawn from the center of the Earth, m 1 = 6 1024 (kg) - mass of the Earth, m - mass of the body, γm 1 =

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

identities r r = r 2 ,

γ m1 m

γ m1 m

γ m1 m

γ m1 m

d A = −

r dr = −

dr = d (−

Π(r) = −

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

at infinity is taken equal to zero.

"

The elementary work of a force on displacement (Fig. 3.22) is the scalar product of a force and the elementary displacement of the point of its application:

where a is the angle between the directions of the vectors and

Because then we can write another expression for elementary work:

For elementary work, you can write a few more expressions:

From the formulas of elementary work it follows that this quantity can be positive (angle a is acute), negative (angle a is obtuse) or equal to zero (angle a is straight).

Full work of forces. To determine the total work done by a force on displacement from a point M 0 to M Let's break this movement down into n displacements, each of which in the limit becomes elementary. Then the work of force A:

Where dA k- work for k-th elementary movement.

The written sum is integral and can be replaced by a line integral taken along the curve at the displacement M 0 M. Then

or

where is the moment in time t=0 corresponds to a point M 0 , and the moment in time t– point M.

From the definition of elementary and complete work it follows:

1) the work of the resultant force on any displacement is equal to the algebraic sum of the work of the component forces on this displacement;

2) the work done by forces on a complete displacement is equal to the sum of the work done by the same force on the component displacements into which the entire displacement is divided in any way.

Power of force. The power of a force is the work done per unit of time:

or considering that

Power power is a quantity equal to the scalar product of the force and the speed of the point of its application.

Thus, at constant power, an increase in speed leads to a decrease in force and vice versa. The unit of power is Watt: 1W=1 J/s.

If a force is applied to a body rotating around a fixed axis, then its power is equal to

The power of a pair of forces is determined similarly.

3.3.4.3. Examples of calculating the work of force

Total work of force -

Where h– the height to which the point has descended.

Thus, the work done by gravity is positive when a point descends and negative when a point rises. The work done by gravity does not depend on the shape of the trajectory between points M 0 and M 1 .

Work of linear elastic force. The linear elastic force is the force acting according to Hooke’s law (Fig. 3.24):

where is the radius vector drawn from the equilibrium point, where the force is zero, to the point in question M; With– constant stiffness coefficient.

Work done by a force on displacement from a point M 0 to point M 1 is determined by the formula

Performing integration, we obtain

(3.27)

Rice. 3.25

Using formula (3.27), the work of the linear elastic force of the springs is calculated when moving along any path from the point M 0, in which its initial deformation is equal to exactly M 1, where the deformation is respectively equal to In the new notation, formula (3.27) takes the form

Work done by a force applied to a rotating rigid body. When a rigid body rotates around a fixed axis, the speed of the point M can be calculated using Euler's formula, see fig. 3.25:

Then we determine the elementary work of force by the formula

Using the mixed cross product property
we get

Because – moment of force relative to a point ABOUT. Considering that – moment of force relative to the axis of rotation Oz and ω dt=dφ, we finally get:

dA=M z dφ.

The elementary work of a force applied to any point of a body rotating around a fixed axis is equal to the product of the moment of force relative to the axis of rotation and the differential of the angle of rotation of the body.

Full work:

In the special case when , the work is determined by the formula

where j is the angle of rotation of the body at which the work of force is calculated.

Rice. 3.26

Work of internal forces of a rigid body. Let us prove that the work done by the internal forces of a rigid body is zero for any movement. It is enough to prove that the sum of the elementary works of all internal forces is equal to zero. Consider any two points of the body M 1 and M 2 (Fig. 3.26). Since internal forces are forces of interaction between points of the body, then:

Let us introduce a unit vector directed along the force. Then

The sum of elementary works of forces and is equal to

Expanding the scalar products of vectors in brackets, we get

Since it has been proven in kinematics that the projections of the velocities of any two points of a rigid body onto the direction of the straight line connecting these points are equal to each other for any motion of the rigid body, then in the resulting expression the difference of identical values ​​is in parentheses, i.e. value equal to zero.

3.3.4.4. Theorem on the change in kinetic energy of a point

For a material point with mass m, moving under the influence of a force, the basic law of dynamics can be represented as

Multiplying both sides of this relation scalarly by the differential of the radius vector of the point we have

or

Considering that – elementary work of force,

(3.28)

Formula (3.28) expresses the theorem on the change in kinetic energy for a point in differential form.

The differential of the kinetic energy of a point is equal to the elementary work of the force acting on the point.

If both sides of equality (3.28) are integrated from the point M 0 to point M(see Fig. 3.22), we obtain a theorem about the change in the kinetic energy of a point in final form:

The change in the kinetic energy of a point at any displacement is equal to the work of the force acting on the point at the same displacement.

3.4.4.5. Theorem on the change in kinetic energy of a system

For each point of the system, the theorem on the change in kinetic energy can be expressed in the form:

Summing the right and left parts of these relations over all points of the system and moving the differential sign beyond the sum sign, we obtain:

or

Where – kinetic energy of the system; – elementary work of external and internal forces, respectively.

Formula (3.29) expresses the theorem about the change in the kinetic energy of the system in differential form.

The differential from the kinetic energy of the system is equal to the sum of the elementary works of all external and internal forces acting on the system.

If both sides of (3.29) are integrated between two positions of the system - initial and final, in which the kinetic energy is equal to T 0 and T, then, changing the order of summation and integration, we have:

or

Where – work of external force for a point in the system Mk when it moves from the initial position to the final position Mk; – work of internal force acting on a point Mk.

Formula (3.30) expresses the theorem about the change in the kinetic energy of the system in finite or integral form.

The change in the kinetic energy of a system when it moves from one position to another is equal to the sum of the work done by all external and internal forces acting on the system on the corresponding movements of points of the system during the same movement of the system.

The work of forces is calculated using the formulas obtained in § 87 and 88. Let us additionally consider the following cases.

1. The work of gravity forces acting on the system. The work of gravity acting on a particle with weight will be equal to where are the coordinates that determine the initial and final positions of the particle (see § 88). Then, taking into account that (see § 32), we find for the sum of the work of all gravity forces acting on the system, the value

This result can also be represented in the form

where P is the weight of the system, is the vertical movement of the center of mass (or center of gravity). Consequently, the work of gravity forces acting on the system is calculated as the work of their main vector (in the case of a rigid body, the resultant) P on the displacement of the center of mass of the system (or the center of gravity of the body).

2. Work of forces applied to a rotating body. The elementary work of the force F applied to the body (Fig. 307) will be equal to (see § 87)

since , where is the elementary angle of rotation of the body.

But, as is easy to see,

We will call the quantity torque. Then we get

Consequently, in the case under consideration, the elementary work is equal to the product of the torque and the elementary angle of rotation. Formula (46) is also valid under the action of several forces, if we assume

When turning to the final angle, work

and in the case of a constant moment

If a body is acted upon by a pair of forces lying in a plane perpendicular to the Oz axis, then in formulas (46)-(47) it will obviously mean the moment of this pair.

Let us also indicate how power is determined in this case (see § 87). Using equality (46), we find

Consequently, when forces act on a rotating body, the power is equal to the product of the torque and the angular velocity of the body. At the same power, the greater the torque, the lower the angular velocity.

3. The work of friction forces acting on a rolling body. A wheel of radius R (Fig. 308), rolling along a certain plane (surface) without sliding, is acted upon by a friction force applied at point B, which prevents the point from sliding along the plane. The elementary work of this force. But point B in this case coincides with the instantaneous center of velocities (see § 56) and

Since then for each elementary movement .

Consequently, when rolling without sliding, the work done by the frictional force that prevents sliding during any movement of the body is zero. For the same reason, in this case the work of the normal reaction N is also zero, if we consider the bodies to be non-deformable due to the force N applied at point B (as in Fig. 308, a).

Let's consider formulas for determining the work and power of a force applied at any point of a rigid body undergoing translational or rotational motion.

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

Let us consider a rigid body undergoing translational motion relative to an inertial reference frame under the influence of a force applied at an arbitrary point (Fig. 24).

In the case of translational motion of a rigid body, all its points move with velocities equal in magnitude and direction. Let us denote the speed of the body.

Using formula (4.31), we obtain

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

Rice. 24. Translational motion of a rigid body under the influence of force

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

where is the angle between the velocity force vectors.

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

Integrating (4.49) on any finite displacement of the point M from the starting position M 0 to position M 1, we get the total work done by the force acting on the body at this displacement

2. Work and power of a force applied to a rigid body undergoing rotational motion.

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

Rice. 25. Rotation of a rigid body around a fixed axis

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

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

Using formula (4.32), we obtain

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

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

The angle between the vectors of torque and angular velocity.

Considering that:

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

2. and therefore

we'll finally get it

Thus, 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 and the differential of the angle of rotation of the body.

To determine the total work done by a force when rotating a body through an angle φ, integrating expression (4.53), we obtain

In the case when , the total 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 angular velocity coincide, then the work done by 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

That is the power of the force applied to a rotating solid body is defined as the product of the moment of force relative to the axis of rotation and the angular velocity of the body . The sign of power is determined similarly to the sign of work.

Theorem: the work done by gravity does not depend on the type of trajectory and is equal to the product of the force modulus and the vertical displacement of the point of its application .

Let the material point M moves under the influence of gravity G and over a certain period of time moves from the position M 1 to position M 2 , having walked the path s (Fig. 4).
On the trajectory of a point M select an infinitesimal area ds , which can be considered rectilinear, and from its ends we draw straight lines parallel to the coordinate axes, one of which is vertical and the other horizontal.
From the shaded triangle we get that

dy = ds cos α.

Elementary work of force G on a way ds is equal to:

dW = F ds cos α.

Total work done by gravity G on a way s equal to

W = ∫ Gds cos α = ∫ Gdy = G ∫ dy = Gh.

So, the work done by gravity is equal to the product of the force and the vertical displacement of the point of its application:

The theorem has been proven.

An example of solving the problem of determining the work of gravity

Task: Homogeneous rectangular array ABCD mass m = 4080 kg has the dimensions indicated on rice. 5.
Determine the work required to tilt the array around an edge D .

Solution.
Obviously, the required work will be equal to the work of resistance performed by the force of gravity of the array, while the vertical movement of the center of gravity of the array when tipping over an edge D is the path that determines the amount of work done by gravity.

First, let's determine the gravity of the array: G = mg = 4080×9.81 = 40,000 N = 40 kN.

To determine vertical movement h center of gravity of a rectangular homogeneous array (it is located at the point of intersection of the diagonals of the rectangle), we use the Pythagorean theorem, based on which:

KO 1 = ОD – КD = √(ОК 2 + КD 2) – КD = √(3 2 +4 2) - 4 = 1 m.

Based on the theorem on the work of gravity, we determine the required work required to overturn the massif:

W = G×KO 1 = 40,000×1 = 40,000 J = 40 kJ.

The problem is solved.

Work done by a constant force applied to a rotating body

Let's imagine a disk rotating around a fixed axis under the influence of a constant force F (Fig. 6), the application point of which moves with the disk. Let's break down the power F into three mutually perpendicular components: F 1 – circumferential force, F 2 – axial force, F 3 – radial force.

When rotating the disk through an infinitesimal angle dφ force F will perform elementary work, which, based on the resultant work theorem, will be equal to the sum of the work of the components.

It is obvious that the work of the components F 2 And F 3 will be equal to zero, since the vectors of these forces are perpendicular to the infinitesimal displacement ds application points M , therefore the elementary work of force F equal to the work of its component F 1 :

dW = F 1 ds = F 1 Rdφ.

When turning the disk to its final angle φ work of force F equal to

W = ∫ F 1 Rdφ = F 1 R ∫ dφ = F 1 Rφ,

where is the angle φ expressed in radians.

Since the moments of the components F 2 And F 3 relative to the axis z are equal to zero, then, based on Varignon’s theorem, the moment of force F relative to the axis z equal to:

M z (F) = F 1 R.

The moment of force applied to the disk relative to the axis of rotation is called torque, and, according to the standard ISO, denoted by the letter T :

T = M z (F), hence, W = Tφ .

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

Example of problem solution

Task: a worker rotates the winch handle by force F = 200 N, perpendicular to the radius of rotation.
Find work spent during time t = 25 seconds, if the length of the handle r = 0.4 m, and its angular velocity ω = π/3 rad/s.

Solution.
First of all, let's determine the angular displacement φ winch handles for 25 seconds:

φ = ωt = (π/3)×25 = 26.18 rad.

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

Power

The work done by any force can be done over different periods of time, that is, at different speeds. To characterize how quickly work is done, in mechanics there is a concept power , which is usually denoted by the letter P .