Theorem on the change in the kinetic energy of the system. Kinetic energy change theorem Energy change theorem

An example of solving a problem using the theorem on the change in the kinetic energy of a system with rigid bodies, blocks, pulleys and a spring.

Content

The task

The mechanical system consists of weights 1 and 2, stepped pulley 3 with step radii R 3 \u003d 0.3 m, r 3 \u003d 0.1 m and radius of gyration relative to the axis of rotation ρ 3 = 0.2 m, block 4 of radius R 4 = 0.2 m and moving block 5. Block 5 is considered to be a continuous homogeneous cylinder. The coefficient of friction of the load 2 about the plane f = 0,1 . The bodies of the system are connected to each other by threads thrown over blocks and wound on pulley 3. Sections of the threads are parallel to the corresponding planes. A spring is attached to the movable block 5 with a stiffness coefficient c = 280 N/m.

Under the action of a force F = f (s) = 80(6 + 7 s) N, depending on the displacement s of the point of its application, the system comes into motion from a state of rest. The deformation of the spring at the moment of the beginning of the movement is equal to zero. When moving, pulley 3 is subjected to a constant moment M = 1.6 Nm resistance forces (from friction in bearings). Masses of bodies: m 1 = 0 , m 2 = 5 kg, m 3 = 6 kg, m 4 = 0 , m 5 = 4 kg.

Determine the value of the center of mass of the body 5 V C 5 at the moment when the displacement s of load 1 becomes equal to s 1 = 0.2 m.

indication. When solving a problem, use theorem on the change in kinetic energy.

The solution of the problem

Given: R 3 \u003d 0.3 m, r 3 \u003d 0.1 m, ρ 3 = 0.2 m, R 4 = 0.2 m, f = 0,1 , s = 280 N/m, m 1 = 0 , m 2 = 5 kg, m 3 = 6 kg, m 4 = 0 , m 5 = 4 kg, F = f (s) = 80(6 + 7 s) N, s 1 = 0.2 m.

Find: VC 5 .

Variable notation

R 3 , r 3- radii of the steps of the pulley 3;
ρ 3 - the radius of inertia of the pulley 3 relative to the axis of rotation;
R 5 - block radius 5;
V 1 , V 2 - speeds of bodies 1 and 2;
ω 3 - angular speed of rotation of the pulley 3;
VC 5 - speed of the center of mass C 5 block 5;
ω 5 - angular speed of rotation of block 5;
s 1 , s 2 - movement of bodies 1 and 2;
φ 3 - angle of rotation of the pulley 3;
s C 5 - displacement of the center of mass C 5 block 5;
s A , s B - displacement of points A and B.

Establishment of kinematic relationships

Let us establish kinematic relations. Since weights 1 and 2 are connected by one thread, their speeds are equal:
V 2 = V1.
Since the thread connecting weights 1 and 2 is wound on the outer step of pulley 3, the points of the outer step of pulley 3 move at a speed V 2 = V1. Then the angular speed of rotation of the pulley:
.
Center of mass velocity V C 5 block 5 is equal to the speed of the points of the inner stage of pulley 3:
.
The speed of point K is zero. Therefore, it is the instantaneous center of velocities of block 5. The angular velocity of rotation of block 5:
.
The speed of point B - the free end of the spring - is equal to the speed of point A:
.

Let us express the velocities in terms of V C 5 .
;
;
.

Now let's install relationships between body movements and rotation angles pulley and block. Since velocities and angular velocities are time derivatives of displacements and angles of rotation
,
then the same connections will be between displacements and rotation angles:
s 2 = s1;
;
;
.

Determining the kinetic energy of a system

Let us find the kinetic energy of the system. Load 2 moves forward at a speed V 2 . Pulley 3 rotates with an angular velocity of rotation ω 3 . Block 5 performs a plane-parallel movement. It rotates with an angular velocity ω 5 and its center of mass moves at a speed V C 5 . Kinetic energy of the system:
.

Since the radius of gyration of the pulley relative to the axis of rotation is given, the moment of inertia of the pulley relative to the axis of rotation is determined by the formula:
J 3 = m 3 ρ 2 3.
Since block 5 is a solid homogeneous cylinder, its moment of inertia about the center of mass is
.

Using kinematic relations, we express all velocities in terms of V C 5 and substitute the expressions for the moments of inertia into the formula for the kinetic energy.
,
where we introduced the constant
kg.

So, we have found the dependence of the kinetic energy of the system on the velocity of the center of mass V C 5 moving block:
, where m = 75 kg.

Determination of the sum of work of external forces

Consider external forces acting on the system.
In this case, we do not consider the tension forces of the threads, since the threads are inextensible and, therefore, they do not produce work. For this reason, we do not consider internal stresses acting in bodies, since they are absolutely rigid.
A given force F acts on body 1 (with zero mass).
Gravity force P acts on load 2 2 = m 2 g 2 and friction force F T .
Pulley 3 is affected by gravity P 3 = m 3 g, pressure force of the N axis 3 and moment of friction force M .
Pulley 4 (with zero mass) is subjected to the pressure force of the axis N 4 .
The movable block 5 is affected by gravity P 5 = m 5 g, spring force F y and thread tension force T K at point K .

The work that the force does when moving the point of its application to a small displacement is equal to the scalar product of the vectors, that is, the product of the modules of the vectors F and ds and the cosine of the angle between them. The given force applied to body 1 is parallel to the movement of body 1. Therefore, the work done by force when moving body 1 at a distance s 1 is equal to:


J.

Consider load 2. It is affected by gravity P 2 , surface pressure force N 2 , thread tension forces T 23 , T 24 and friction force F T . Since the load does not move in the vertical direction, the projection of its acceleration on the vertical axis is zero. Therefore, the sum of the projections of forces on the vertical axis is zero:
N 2 - P 2 = 0;
N 2 \u003d P 2 \u003d m 2 g.
Friction force:
F T = fN 2 = f m 2 g.
P-forces 2 and N 2 perpendicular to the displacement s 2 so they don't do any work.
The work of the friction force:
J.

If we consider load 2 as an isolated system, then we need to take into account the work done by the tension forces of the threads T 23 and T 24 . However, we are interested in the entire system consisting of bodies 1, 2, 3, 4 and 5. For such a system, the thread tension forces are internal forces. And since the threads are inextensible, the sum of their work is zero. In the case of load 2, it is also necessary to take into account the tension forces of the threads acting on pulley 3 and block 4. They are equal in magnitude and opposite in direction to the forces T 23 and T 24 . Therefore, the work done by the tension forces of the threads 23 and 24 over the load 2 is equal in magnitude and opposite in sign to the work done by the tension forces of these threads over the pulley 3 and block 4. As a result, the sum of the work done by the tension forces of the threads is zero.

Consider pulley 3. Since its center of mass does not move, the work of gravity P 3 equals zero.
Since the C-axis 3 is stationary, then the pressure force of the N axis 3 does not produce work.
The work produced by the moment of forces is calculated similarly to the work produced by the force:
.
In our case, the vectors of the moment of friction forces and the angle of rotation of the pulley are directed along the axis of rotation of the pulley, but opposite in direction. Therefore, the work of the moment of friction forces:
J.

Consider block 5.
Since the speed of point K is equal to zero, then the force T K does not produce work.
Center of gravity of block C 5 moved a distance s C 5 up. Therefore, the work done by the gravity of the block is:
J.
The work of the elastic force of the spring is equal to the change in the potential energy of the spring with a minus sign. Since the spring is not deformed at first, then
J.

The sum of the work of all forces:

J.

Application of the theorem on the change in the kinetic energy of the system

We apply the theorem on the change in the kinetic energy of the system in integral form.
.
Since the system was at rest at the beginning, its kinetic energy at the beginning of the motion
T 0 = 0 .
Then
.
From here
m/s.

Lecture 5 Kinetic energy change theorem

5. 1. Work of force

Let the power is the resultant of all forces of the system, applied to the point Р, and ( dx, dy, dz) - elementary movement of the point P along its trajectory P 1 P 2 (Fig. 5.1). elementary work dA forces is called the scalar product

Elementary work is a scalar quantity. If is the angle between the force and the direction of displacement , then expression (5.1) can be represented as

where is the projection of the force on the direction of the elementary displacement (or the direction of the speed of the point).

The sign of the elementary work depends on the sign of the function . If is an acute angle, then , if is an obtuse angle, then , if , then .

Let the point R makes a final move from position to position , describing an arc . Let's split the arc into n arbitrary small sections, indicating the length of the section with the number k through . Then the elementary work of the force on k-th section will be equal to , and all the way from to - the sum of work on individual sections

We obtain the exact value of the work by passing to the limit, provided that the number of sections n increases indefinitely, and the length of each section decreases:

.

Such a limit is called a curvilinear integral of the first kind over an arc and is written as follows

. (5.3)

The result of the integration is the complete work A strength F on the considered finite displacement along the path .

5. 1. 1. Work of gravity

Let m is the mass of the point, g- acceleration of gravity. Then

Calculating the work by formulas (5.1) and (5.3), we have

where is the drop height.

When the point is raised, therefore, .

5. 1. 2. Work of the linear force of elasticity

Let the material point R moves along the axis Oh(Fig. 5.3) under the action of the spring to which it is attached. If at , , then the spring is deformed, and with small deviations of the point, we can assume that an elastic force is applied to it from the side of the spring. Then the work of the elastic force on displacement x 0 x 1 will be equal to

. (5.5)

The work of the elastic force is equal to half the product of the stiffness coefficient and the difference between the squares of the initial and final elongation (or compression) of the spring.

5. 1. 3. Elementary work of forces applied to a rigid body

Consider the motion of a body in a plane. Let ABOUT- an arbitrarily chosen point on a solid body (Fig. 5.4). Let's call it a pole. Then the motion of the body in a plane can be represented as the sum of the simplest: translational motion along with the pole and rotation of the body around the pole. Then, the speed of a point relative to a fixed coordinate system is defined as the geometric sum of two speeds

where is the velocity of the pole, is the vector of the angular velocity of the rigid body, is the Euler velocity, i.e. the velocity of the point as it rotates around the pole.

We will represent a rigid body as a mechanical system consisting of N individual points, the mutual distance between which does not change.

Calculate the displacement of a point under the action of a force:

Then .

Elementary work, according to (5.1), can be written as follows

Using the properties of the mixed product of vectors , we rewrite the last expression in the form

Let - the resultant of all forces, external and internal (Fig. 5.4), applied at a point of the body, i.e.

.

Then (a) is written as

According to (3.1 and 3.2), the main vector and the main moment of the internal forces of the system are equal to zero, we obtain

Here: is the main vector, is the main moment of external forces about the point ABOUT.

Special cases

A. Translational motion of a rigid body. All points of the body have the same displacements (Fig. 5.5, a) both in absolute value and in direction, then, from (5.6), we obtain (here):

. (5.7)

b. Rotation of a rigid body about a fixed axis. Let the axis z passes through the pole ABOUT(Fig. 5.5b). Then , ; from (5.6) we get

. (5.8)

Example. coil mass m and radius R propelled by a constant force F, applied at the point A(Fig. 5.6). The coil rolls to the right without slipping on a rough surface.

Calculate the work of all external forces if the center of the coil has moved a distance , is the coefficient of rolling friction, is the friction force, r is the radius of the core of the coil to which the force is applied.

Solution. The coil makes a flat motion. Since the rolling occurs without sliding, the instantaneous center of velocities is located at the point of contact of the coil with the plane, i.e. at the point R(fig.5.6). Let's direct the S axis horizontally to the right. In accordance with the direction of movement, we take the positive direction of the angle of rotation counterclockwise.

Let the center of the coil WITH will move to . In this case, the coil will turn by an angle . Then from where

Taking a point R for the instantaneous axis of rotation, we calculate the elementary work using the formula (5.8):

(A)

Here: lines of action of forces and mg cross the axis of rotation, so ; further , where N is the normal reaction force.

To determine the desired work, it remains to take a definite integral of (a) in the range from 0 to SA. Get

5. 2. Force field. Power function. Potential energy

Suppose that a point is moving in some space and a force acts on it from the side of space, which depends on the position of the point in this space, but does not depend on the speed of the point. In this case, we say that the space is given force field, and also that the point moves in the force field. The corresponding concepts for a system of material points are similar.

Forces that depend on the position of their points of application are often encountered in mechanics. For example, the elastic force applied to a material point that moves along a horizontal straight line under the action of a spring. The most important example of a force field in nature is the gravitational field: the action of the Sun on a planet of a given mass is determined at every point in space by the law of universal gravitation.

The force field is called potential if there is a scalar function U, depending only on the coordinates , , point-point of the material system (possibly also on time), such that

The function is called power function.

Consider the properties of the strength function.

Elementary work (5.1) is related to the force function as follows

Thus, the elementary work of a force in a potential force field is equal to the total differential of the force function ii.

The total work of the force on the section from the point to the point (fig.5.1)

those. . (5.10)

It follows from the obtained expressions that

1. the work of a force in a potential force field along any closed path is zero;

2. the work of a force in a potential force field depends only on the position of the final and initial points, but the path itself does not play a role.

Potential energy. Potential energy P at the considered point of the force field R called the work done by the field forces acting on a material point when it moves from a point R to the starting point 1 , i.e.

P= or P=

Let's connect the force function U with potential energy. We have

Examples of calculating potential energy

1. Uniform gravity field. Let m is the mass of the point; g - acceleration of gravity. Then (Fig. 5.2)

2. Force field of an elastic spring. Let the material point move along the axis Oh(Fig. 5.3) under the action of the spring to which it is attached. If at , the spring is not deformed, then, assuming in formula (5.5) , we obtain

.

5. 3. Kinetic energy

5. 3. 1. Kinetic energy of the system. König's theorem

The kinetic energy of a material point is called half the product of the mass of the point and the square of its speed, i.e. . Kinetic energy is a positive scalar quantity. In the SI system, the unit of kinetic energy is the joule: .

The kinetic energy of a mechanical system is the sum of the kinetic energies of all points included in the system:

(5.11)

The velocities of the points of the system (5.1) are determined with respect to the fixed frame of reference.

Let's combine the origin of coordinates with the center of mass of the system. Suppose that the mechanical system, together with the coordinate system, moves translationally relative to the fixed coordinate system (Fig. 5.7). Point - point of the system.

Then, based on the velocity addition theorem, the absolute velocity of the point Rk. system is written as the vector sum of the translational and relative velocities:

, (A)

where is the speed of the origin of the moving coordinate system (transfer speed, i.e. the speed of the center of mass of the system); - point speed Rk relative to the moving coordinate system Ohuz (relative speed).

Substituting (a) into formula (5.11), we obtain

(5.12)

Here is the mass of the entire system.

The radius vector of the center of mass of the system in the moving coordinate system is determined, according to (2.1), - , where , i.e. . Because the origin ABOUT is the center of mass of the system, then , then , i.e. the second sum in expression (5.12) is equal to zero.

Thus, the kinetic energy of system (5.12) has the form

(5.13)

This equality defines Koenig's theorem.

Theorem. The kinetic energy of the system is equal to the sum of the kinetic energy that a material point located at the center of mass of the system and having a mass equal to the mass of the system and the kinetic energy of the system's motion relative to the center of mass would have.

5. 3. 2. Kinetic energy of a solid body

A solid body is a special case of a mechanical system and is considered as a continuously distributed mass, then all the sums included in the expression for the kinetic energy of the system go into integrals. Thus, for a rigid body formula (5.11) takes the form

. (5.14)

1. Kinetic energy of a rigid body moving forward.

With this type of movement, the speeds of all points of the body are the same (Fig. 5.8). Taking it out of the integral sign in formula (5.14), we obtain

. (5.15)

The kinetic energy of a rigid body moving forward is equal to half the product of the mass of the bodyMto the square of its speed.

2. Kinetic energy of a rigid body rotating about a fixed axis

Speed ​​module V any point of a rigid body rotating around a fixed axis is equal to , where is the modulus of the angular velocity of the rigid body, is the distance from the point to the axis of rotation z(Fig. 5.9). Substituting into formula (5.14), we obtain

Here is the moment of inertia of the rigid body about the axis z.

The kinetic energy of a rigid body rotating around a fixed axis is equal to half the product of the moment of inertia of the body about the axis of rotation and the square of the angular velocity of the body.

3. Kinetic energy of a rigid body during plane-parallel motion

With plane-parallel motion, the speed of any point of the body consists of the geometric sum of the speed of the pole and the speed of the point during rotation around the pole. Let the body move flat in a plane Oxy, Then

|| . For the pole we choose the center of mass of the body, then in the formula (5.13), the speed is the speed of the point k body during its rotation relative to the pole (center of mass) and is equal to , where the distance k- oh point to the pole. Then (5.13) is rewritten

Bearing in mind that is the moment of inertia of the body about the axis z passing through the pole WITH, the last expression can be rewritten as

, (5.17)

with a plane-parallel motion of the body, the kinetic energy is the sum of the kinetic energy of translational motion together with the center of mass and the kinetic energy from rotation around an axis passing through the center of mass and perpendicular to the plane of motion.

5. 4. Theorem on the change in kinetic energy

5. 4. 1. Theorem on the change in the kinetic energy of a point

Find the relationship between work and change in speed. Let a material point with mass m moves along the axis Oh under the action of a force, for example, a compressed or unclenched spring, fixed at the origin, - a point ABOUT(Fig. 5.10). The equation of motion of a point has the form

Multiply both sides of this equation by , and, given that , we get

. (5.19)

On the right side of this equality, we replace Vx by and multiply by dt right and left parts. Then

. (5.20)

In this form, the equality has a very clear meaning: when the point is shifted by dx, the force does work , as a result of which the value changes point kinetic energy characterizing the motion of a point and, in particular, the modulus of its velocity. If the point is displaced from position in , and its speed changes from to , then, integrating (5.20), we have

. (5.21)

Given that , we finally find

. (5.22)

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

Doing all the previous procedures, we get

,

here is the arc along which the point moves (Fig. 5.11).

5. 4. 2. Theorem on the change in the kinetic energy of the system

Let the points of the system by mass move in such a way that their radius vectors in the inertial frame of reference have received an increment . Let us find how the kinetic energy has changed in this case T systems.

According to (5.11), the kinetic energy of the system

.

Let us calculate the differential of the kinetic energy of the system and transform the resulting expression

Here

Taking into account that , where is the acceleration of the point a and are the resultants of external and internal forces applied to the point, we rewrite the last equality in the form

Thus,

. (5.23)

The last equality expresses the theorem on the change in the kinetic energy of a mechanical system in differential form: the differential of the kinetic energy of the system is equal to the elementary work of all the forces of the system.

special case. For an absolutely rigid body, the sum of the work of all internal forces of the system is zero:

.

Consequently, the theorem on the change in kinetic energy (5.23) for a rigid body can be written as

The change in the kinetic energy of a rigid body with any elementary displacement is equal to the elementary work of external forces acting on the body.

If both parts of (5.24) are integrated between two positions - initial and final, in which, respectively, the kinetic energy and , we obtain

. (5.25)

Example 1. Disk mass m=5 kg and the radius is driven by a constant force applied at a point A(Fig. 5.6). A disk rolls on a rough surface to the right without slipping. Determine the speed of the center of mass WITH coil at the moment when it moves a distance , coefficient of sliding friction , , radius of gyration of the disk

Solution. The disc is moving in a plane. Let us write down the theorem on the change in kinetic energy for a rigid body

Let us calculate the kinetic energy of the disk. At the initial moment of time, the disk was at rest, i.e. . Kinetic energy in the final position of the disc

If we consider any point of the system with mass , having speed , then for this point will be

,

where and - elementary work of external and internal forces acting on a point. Compiling such equations for each of the points of the system and adding them term by term, we obtain

,

. (2)

The equality expresses the theorem on the change in the kinetic energy of the system in differential form.

If the resulting expression is attributed to the elementary period of time during which the movement under consideration occurred, we can obtain the second formulation for the differential form of the theorem: the time derivative of the kinetic energy of the mechanical system is equal to the sum of the powers of all external () and internal () forces, i.e.

Differential forms of the theorem on the change in kinetic energy can be used to compile differential equations of motion, but this is done quite rarely, because there are more convenient methods.

Having integrated both parts of equality (2) within the limits corresponding to the displacement of the system from some initial position, where the kinetic energy is , to a position where the value of the kinetic energy becomes equal to , will have

The resulting equation expresses the theorem on the change in kinetic energy in the final form: the change in the kinetic energy of the system during some of its displacement is equal to the sum of the work on this displacement of all external and internal forces applied to the system.

Unlike the previous theorems, internal forces in the equations are not excluded. Indeed, if and are the forces of interaction between the points and the system (see Fig. 51), then . But at the same time, the point , can move towards , and the point can move towards . The work of each of the forces will then be positive and the sum of the work will not be zero. An example is the phenomenon of rollback. Internal forces (pressure forces) acting both on the projectile and on the rolling parts do positive work here. The sum of these works, which is not equal to zero, changes the kinetic energy of the system from the value at the beginning of the shot to the value at the end.

Another example: two points connected by a spring. When the distance between points changes, the elastic forces applied to the points will do work. But if the system consists of absolutely rigid bodies and the connections between them are unchangeable, not elastic, ideal, then the work of internal forces will be equal to zero and they can be ignored and not shown at all on the calculation diagram.

Let us consider two important special cases.

1) Immutable system. immutable we will call a system in which the distances between the points of application of internal forces do not change during the motion of the system. In particular, such a system is an absolutely rigid body or an inextensible thread.

Fig.51

Let two points and of an unchanging system (Fig. 51) acting on each other with forces and () have speeds and at a given moment. Then for a period of time dt these points will make elementary displacements and , directed along the vectors and . But since the segment is unchangeable, then, according to the well-known theorem of kinematics, the projections of the vectors and , and, consequently, both displacements and the direction of the segment will be equal to each other, i.e. . Then the elementary work of the forces and will be identical in absolute value and opposite in sign and will add up to zero. This result is valid for all internal forces for any displacement of the system.

Hence we conclude that for an unchanging system, the sum of the work of all internal forces is zero and the equations take the form

2) A system with ideal connections. Let us consider a system on which constraints are imposed that do not change with time. We divide all external and internal forces acting on the points of the system by active And bond reactions. Then

,

where is the elementary work of acting on k- the th point of the system of external and internal active forces, a is the elementary work of the reactions of external and internal bonds imposed on the same point.

As you can see, the change in the kinetic energy of the system depends on the work and active forces and reactions of the bonds. However, it is possible to introduce the concept of such "ideal" mechanical systems, in which the presence of bonds does not affect the change in the kinetic energy of the system during its motion. For such connections, obviously, the following condition must be satisfied:

If for bonds that do not change with time, the sum of the work of all reactions during an elementary displacement of the system is equal to zero, then such bonds are called ideal. For a mechanical system, on which only ideal constraints that do not change with time are imposed, we will obviously have

Thus, the change in the kinetic energy of a system with ideal bonds that do not change with time during any of its displacements is equal to the sum of the work on this displacement applied to the system of external and internal active forces.

The mechanical system is called conservative(its energy is, as it were, conserved, does not change), if the integral of energy takes place for it

or (3)

It is the law of conservation of mechanical energy: when a system moves in a potential field, its mechanical energy (the sum of potential and kinetic) remains unchanged, constant all the time.

A mechanical system will be conservative if the forces acting on it are potential, for example, gravity, elastic forces. In conservative mechanical systems, using the energy integral, one can check the correctness of the formulation of differential equations of motion. If the system is conservative, and condition (3) is not satisfied, then an error was made when compiling the equations of motion.

The energy integral can be used to check the correctness of the equations in another way, without calculating the derivative. To do this, after carrying out the numerical integration of the equations of motion, calculate the value of the total mechanical energy for two different moments of time, for example, initial and final. If the difference between the values ​​turns out to be comparable with the calculation errors, this will indicate the correctness of the equations used.

All previous theorems made it possible to exclude internal forces from the equations of motion, but all external forces, including the previously unknown reactions of external constraints, were preserved in the equations. The practical value of the theorem on the change in kinetic energy lies in the fact that, with ideal connections that do not change with time, it will allow us to exclude from the equations of motion All previously unknown reactions of bonds.

mechanical system

The kinetic energy of a mechanical system is called the arithmetic sum of the kinetic energies of all its material points

Calculation of the kinetic energy of a rigid body

1. Forward movement

As is known, in the case of translational motion, the velocities of all points of the body at the same moment of time are equal, then (83) can be represented as

. (84)

In the translational motion of a body, its kinetic energy is equal to half the product of the mass and the square of the velocity of the center of mass.

2. Rotational motion of a rigid body

P In rotational motion, the speed of each point of the body

. (85)

Substitute (85) into (83):

.

Taking into account (59), we obtain

. (86)

During rotational motion, the kinetic energy is equal to half the product of the moment of inertia of the body about the axis of rotation and the square of the angular velocity.

3 . flat motion

Plane motion can be represented as rotation about the pole (for example, the center of mass) and movement along with the pole, then

. (87)

The kinetic energy of a body in plane motion is equal to the sum of the kinetic energies from translational motion along with the center of mass and rotational motion relative to the center of mass.

Theorem: The change in the kinetic energy of a mechanical system at some of its displacement is equal to the sum of the work of all internal and external forces of the system at the same displacement

. (88)

Remarks:

1. The introduced value of the kinetic energy of the system, in contrast to the momentum of the system and the angular momentum, is a scalar quantity. Wherein:

Q=0 during rotation and rest;

K O=0 in translational motion or rest;

T

Thus, in contrast to the theorem on the change in momentum and angular momentum, this theorem is suitable for studying any kind of motion, since T=0 only for a stationary system.

2. Unlike the mentioned theorems, this theorem takes into account the action of the internal forces of the system.

Some cases of work calculation

1. The work of the moment of forceM Z relative to the axis is equal to the product of the moment and the angle of rotation  bodies about the axis

. (89)

2. The sum of the work of internal forces absolutely rigid body (non-deformable) is always equal to zero.

3. The work of the rolling friction moment
.

,

Where  - coefficient of rolling friction;

R is the radius of the cylinder;

s is the length of the arc, equal to the segment of the path traveled by the center of mass C along the surface;


- the angle of rotation of the axes of the cylinder in the process of movement;

N– normal surface reaction;

P- gravity;

F tr is the force of sliding friction.

Differential equations of translational, rotational and plane motion of a rigid body

1. translational movement

In translational motion, all points of the body move along the same trajectories and at the same time have the same acceleration. Then the motion of the center of mass theorem (67) can be used to describe the motion. We project this equation onto the coordinate axes

System (90) is a differential equation for the translational motion of a rigid body.

2. rotational movement

P Let a rigid body rotate about an axis under the action of forces. The dynamic characteristic of the rotational motion of a rigid body is the kinetic moment K z, and the characteristic of the rotational action of the force is the moment of force about the axis. Therefore, to describe the rotational motion of a rigid body relative to a fixed axis, we use the theorem on the change in angular momentum (81)

. (91)

When rotating
, Then

,

given that I z=const, as a result we get

. (92)

Equation (92) is a differential equation for the rotational motion of a rigid body around a fixed axis.

found angle  will determine the position of a rotating body at any given time.

3. flat motion

The position of a body making a plane motion at any moment of time is determined by the position of the pole and the angle of rotation of the body relative to the pole. If we take the center of mass of the body as a pole, then the equation of its motion can be found from the theorem on the motion of the center of mass (67), and the rotational motion about the center will be determined by equation (92), which is also valid for the motion of the system about the axis passing through the center of mass. Then the differential equations of plane motion of a rigid body have the form

The kinetic energy of a mechanical system is the sum of the kinetic energies of all its points:

Differentiating each part of this equality with respect to time, we obtain

Using the basic law of dynamics for To-th point of the system m k 2i k= Fj., we arrive at the equality

The scalar product of the force F and the speed v of the point of its application is called power of force and denote R:

Using this new notation, we represent (11.6) in the following form:

The resulting equality expresses the differential form of the theorem on the change in kinetic energy: the rate of change of the kinetic energy of a mechanical system is equal to the sum j of the powers of all cm acting on the system.

Representing the derivative f in (8.5) in the form of a fraction -- and doing

then separation of variables, we get:

Where dT- kinetic energy differential, i.e. its change over an infinitesimal time interval dr, dr k = k dt - elementary displacement To- th point of the system, i.e. movement in time dt.

The scalar product of the force F and the elementary displacement dr its points of application are called elementary work forces and represent dA:

Using the properties of the scalar product, one can represent the elementary work of the force also in the form

Here ds = dr- the length of the arc of the trajectory of the point of application of the force, corresponding to its elementary displacement c/g; A - the angle between the directions of the force vector F and the elementary displacement vector c/r; F „ F y , F,- projections of the force vector F on the Cartesian axes; dx, dy, dz projections onto the Cartesian axes of the elementary displacement vector s/r.

Taking into account the notation (11.9), equality (11.8) can be represented in the following form:

those. the differential of the kinetic energy of the system is equal to the sum of the elementary works of all forces acting on the system. This equality, like (11.7), expresses the differential form of the theorem on the change in kinetic energy, but differs from (11.7) in that it uses not derivatives, but infinitesimal increments - differentials.

Performing term-by-term integration of equality (11.12), we obtain

where as the limits of integration are used: 7 0 - kinetic energy of the system at the moment of time? 0; 7) - kinetic energy of the system at the moment of time t x .

Definite integrals over time or A(F):

Remark 1. To calculate the work, it is sometimes more convenient to use a non-arc parametrization of the trajectory M(s), and the coordinate M(x(t), y(/), z(f)). In this case, for the elementary work, it is natural to take the representation (11.11), and represent the curvilinear integral in the form:

Taking into account the designation (11.14) of work on a finite displacement, equality (11.13) takes the form

and represents the final form of the theorem on the change in the kinetic energy of a mechanical system.

Theorem 3. The change in the kinetic energy of a mechanical system when it moves from the initial position to the final one is equal to the sum of the work of all forces acting on the points of the system during this movement.

Comment 2. The right side of equality (11.16) takes into account the works all the forces acting on the system, both external and internal. Nevertheless, there are such mechanical systems for which the total work of all internal forces is equal to zero. Ego so called immutable systems, for which the distances between interacting material points do not change. For example, a system of rigid bodies connected by frictionless hinges or flexible inextensible threads. For such systems, in equality (11.16) it is sufficient to take into account only the work of external forces, i.e. theorem (11.16) takes the form: