Theorem on the change in momentum of a mechanical system. Theorem on the change in momentum
The system referred to in the theorem can be any mechanical system consisting of any bodies.
Statement of the theorem
The amount of motion (momentum) of a mechanical system is a value equal to the sum of the quantities of motion (momentum) of all bodies included in the system. The impulse of external forces acting on the bodies of the system is the sum of the impulses of all external forces acting on the bodies of the system.
( kg m/s)
The theorem on the change in the momentum of the system states
The change in the momentum of the system over a certain period of time is equal to the impulse of external forces acting on the system over the same period of time.
Law of conservation of momentum of the system
If the sum of all external forces acting on the system is equal to zero, then the momentum (momentum) of the system is a constant value.
,
we obtain the expression of the theorem on the change in the momentum of the system in differential form:
Having integrated both parts of the resulting equality over an arbitrarily taken time interval between some and , we obtain the expression of the theorem on the change in the momentum of the system in integral form:
Law of conservation of momentum (Law of conservation of momentum) states that the vector sum of the impulses of all bodies of the system is a constant value if the vector sum of the external forces acting on the system is equal to zero.
(Moment of momentum m 2 kg s −1)
Theorem on the change in the angular momentum about the center
the time derivative of the moment of momentum (kinetic moment) of a material point relative to any fixed center is equal to the moment of the force acting on the point relative to the same center.
dk 0 /dt = M 0 (F ) .
Theorem on the change in the angular momentum about the axis
the time derivative of the moment of momentum (kinetic moment) of a material point with respect to any fixed axis is equal to the moment of the force acting on this point with respect to the same axis.
dk x /dt = M x (F ); dk y /dt = M y (F ); dk z /dt = M z (F ) .
Consider a material point M weight m moving under the influence of a force F (Figure 3.1). Let's write down and construct the vector of the angular momentum (kinetic momentum) M 0 material point relative to the center O :
Differentiate the expression for moment of momentum (kinetic moment k 0) by time:
Because dr /dt = V , then vector product V ⊗ m ⋅ V (collinear vectors V and m ⋅ V ) is zero. In the same time d(m ⋅ v) /dt = F according to the momentum theorem material point. Therefore, we get that
dk 0 /dt = r ⊗F , (3.3)
where r ⊗F = M 0 (F ) – vector-moment of force F relative to the fixed center O . Vector k 0 ⊥ plane ( r , m ⊗V ), and the vector M 0 (F ) ⊥ plane ( r ,F ), we finally have
dk 0 /dt = M 0 (F ) . (3.4)
Equation (3.4) expresses the theorem on the change in the angular momentum (kinetic momentum) of a material point relative to the center: the time derivative of the moment of momentum (kinetic moment) of a material point relative to any fixed center is equal to the moment of the force acting on the point relative to the same center.
Projecting equality (3.4) onto the axes of Cartesian coordinates, we obtain
dk x /dt = M x (F ); dk y /dt = M y (F ); dk z /dt = M z (F ) . (3.5)
Equalities (3.5) express the theorem on the change in the angular momentum (kinetic moment) of a material point about the axis: the time derivative of the moment of momentum (kinetic moment) of a material point with respect to any fixed axis is equal to the moment of the force acting on this point with respect to the same axis.
Let us consider the consequences following from theorems (3.4) and (3.5).
Consequence 1. Consider the case when the force F during the entire movement of the point passes through the fixed center O (case of central force), i.e. when M 0 (F ) = 0. Then it follows from Theorem (3.4) that k 0 = const ,
those. in the case of a central force, the moment of momentum (kinetic moment) of a material point relative to the center of this force remains constant in magnitude and direction (Figure 3.2).
Figure 3.2
From the condition k 0 = const it follows that the trajectory of the moving point is a plane curve, the plane of which passes through the center of this force.
Consequence 2. Let M z (F ) = 0, i.e. force crosses the axis z or parallel to it. In this case, as can be seen from the third of equations (3.5), k z = const ,
those. if the moment of a force acting on a point with respect to some fixed axle is always zero, then the moment of momentum (kinetic moment) of the point about this axis remains constant.
Proof of the momentum change theorem
Let the system consist of material points with masses and accelerations . All forces acting on the bodies of the system can be divided into two types:
External forces - forces acting from bodies that are not included in the system under consideration. The resultant of external forces acting on a material point with the number i denote .
Internal forces are the forces with which the bodies of the system itself interact with each other. The force with which the point with the number i point number is valid k, we will denote , and the impact force i-th point on k-th point - . Obviously, for , then
Using the introduced notation, we write Newton's second law for each of the considered material points in the form
Given that 
and summing up all the equations of Newton's second law, we get:
The expression is the sum of all internal forces operating in the system. According to Newton's third law, in this sum, each force corresponds to a force such that and, therefore, is fulfilled 
Since the whole sum consists of such pairs, the sum itself is equal to zero. Thus, one can write
Using the designation for the momentum of the system, we obtain
Introducing into consideration the change in the momentum of external forces 
, we obtain the expression of the theorem on the change in the momentum of the system in differential form:
Thus, each of the last obtained equations allows us to assert: the change in the momentum of the system occurs only as a result of the action of external forces, and internal forces cannot have any effect on this value.
Having integrated both parts of the obtained equality over an arbitrarily taken time interval between some and , we obtain the expression of the theorem on the change in the momentum of the system in integral form:
where and are the values of the amount of motion of the system at the moments of time and, respectively, and is the impulse of external forces over a period of time . In accordance with the above and the introduced notation,
Let the material point move under the action of force F. It is required to determine the motion of this point with respect to the moving system Oxyz(see the complex motion of a material point), which moves in a known way with respect to a fixed system O 1 x 1 y 1 z 1 .
The basic equation of dynamics in a stationary system
Let's write down absolute acceleration points by Coriolis theorem
where a abs– absolute acceleration;
a rel– relative acceleration;
a lane– portable acceleration;
a core is the Coriolis acceleration.
Let us rewrite (25) taking into account (26)
Let us introduce the notation 
- portable force of inertia, 
is the Coriolis force of inertia. Then equation (27) takes the form
The basic equation of dynamics for studying relative motion (28) is written in the same way as for absolute motion, only the translational and Coriolis forces of inertia must be added to the forces acting on the point.
General theorems of material point dynamics
When solving many problems, you can use pre-made blanks obtained on the basis of Newton's second law. Such problem solving methods are combined in this section.
Theorem on the change in momentum of a material point
Let us introduce the following dynamic characteristics:
1. Quantity of movement of a material point is a vector quantity equal to the product of the mass of a point and the vector of its velocity
.
(29)
2. Impulse of force
Elemental Force Impulse- a vector quantity equal to the product of the force vector by an elementary time interval
(30).
Then full impulse
.
(31)
At F=const we get S=ft.
The total impulse over a finite period of time can be calculated only in two cases, when the force acting on the point is constant or depends on time. In other cases, it is necessary to express the force as a function of time.
The equality of the dimensions of momentum (29) and momentum (30) makes it possible to establish a quantitative relationship between them.
Consider the motion of a material point M under the action arbitrary force F along an arbitrary path.
O 
UD: 
.
(32)
We separate variables in (32) and integrate
.
(33)
As a result, taking into account (31), we obtain
.
(34)
Equation (34) expresses the following theorem.
Theorem: The change in the momentum of a material point over a certain period of time is equal to the impulse of the force acting on the point over the same time interval.
When solving problems, equation (34) must be projected on the coordinate axes
This theorem is convenient to use when the given and unknown quantities include the mass of a point, its initial and final velocity, forces, and time of motion.
Theorem on the change in the angular momentum of a material point
M 
moment of momentum of a material point relative to the center is equal to the product momentum modulus of a point per shoulder, i.e. shortest distance (perpendicular) from the center to a line coinciding with the velocity vector
,
(36)
.
(37)
The relationship between the moment of force (cause) and the moment of momentum (effect) is established by the following theorem.
Let point M of given mass m moving under the influence of force F.
,
,
,
(38)
.
(39)
Let us calculate the derivative of (39)
.
(40)
Combining (40) and (38), we finally obtain
.
(41)
Equation (41) expresses the following theorem.
Theorem: The time derivative of the angular momentum vector of a material point relative to some center is equal to the moment of the force acting on the point relative to the same center.
When solving problems, equation (41) must be projected on the coordinate axes
In equations (42), the moments of momentum and force are calculated relative to the coordinate axes.
From (41) it follows law of conservation of angular momentum (Kepler's law).
If the moment of force acting on a material point relative to any center is equal to zero, then the angular momentum of the point relative to this center retains its magnitude and direction.
If 
, then 
.
The theorem and the conservation law are used in curvilinear motion problems, especially under the action of central forces.
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Number of movement
Quantity of movement of a material point - a vector quantity equal to the product of the mass of the point and the vector of its velocity.
The unit of momentum is (kg m/s).
Quantity of movement of the mechanical system - a vector quantity equal to the geometric sum (principal vector) of the momentum of a mechanical system equals the product of the mass of the entire system and the speed of its center of mass.
When a body (or system) moves in such a way that its center of mass is stationary, then the momentum of the body is zero (for example, the rotation of the body around a fixed axis passing through the center of mass of the body).
In the case of complex motion, the momentum of the system will not characterize the rotational part of the motion when rotating around the center of mass. That is, the amount of motion characterizes only the translational motion of the system (together with the center of mass).
Impulse of force
The momentum of a force characterizes the action of a force over a certain period of time.
Impulse of force over a finite period of time is defined as the integral sum of the corresponding elementary impulses.
Theorem on the change in momentum of a material point
(in differential form e ):
The time derivative of the momentum of a material point is equal to the geometric sum of the forces acting on the points.
(v integral form ):
The change in the momentum of a material point over a certain period of time is equal to the geometric sum of the impulses of forces applied to the point over this period of time.
Theorem on the change in the momentum of a mechanical system
(in differential form ):
The time derivative of the momentum of the system is equal to the geometric sum of all external forces acting on the system.
(in integral form ):
The change in the amount of motion of the system over a certain period of time is equal to the geometric sum of the impulses of external forces acting on the system over this period of time.
The theorem makes it possible to exclude obviously unknown internal forces from consideration.
The theorem on the change in the momentum of a mechanical system and the theorem on the movement of the center of mass are two different forms one theorem.
Law of conservation of momentum of the system
- If the sum of all external forces acting on the system is equal to zero, then the momentum vector of the system will be constant in direction and modulo.
- If the sum of the projections of all acting external forces on any arbitrary axis is equal to zero, then the projection of the momentum on this axis is a constant value.
conclusions:
- Conservation laws indicate that internal forces cannot change the total momentum of the system.
- The theorem on the change in the momentum of a mechanical system does not characterize the rotational motion of a mechanical system, but only translational.
An example is given: Determine the amount of motion of a disk of a certain mass, if its angular velocity and size are known.
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Application of the kinetic energy conservation theorem
An example of solving the problem of applying the theorem on the conservation of kinetic energy of a mechanical system
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An example of solving the problem of determining the speed and acceleration of a point by given equations movements
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An example of solving the problem of applying the theorem on the change in angular momentum to determine angular velocity a body that rotates around a fixed axis.
Consisting of n material points. Let us single out some point from this system Mj with mass mj. It is known that external and internal forces act on this point.
Apply to a point Mj resultant of all internal forces F j i and the resultant of all external forces F j e(Figure 2.2). For selected material point Mj(as for a free point) we write the theorem on the change in momentum in differential form (2.3):
We write similar equations for all points of the mechanical system (j=1,2,3,…,n).
Figure 2.2
Let's put everything together n equations:
∑d(m j ×V j)/dt = ∑F j e + ∑F j i, (2.9)
d∑(m j ×V j)/dt = ∑ F j e + ∑ F j i. (2.10)
Here ∑mj ×Vj =Q is the momentum of the mechanical system;
∑ F j e = R e is the main vector of all external forces acting on the mechanical system;
∑ F j i = R i =0- the main vector of the internal forces of the system (according to the property of internal forces, it is equal to zero).
Finally, for the mechanical system, we obtain
dQ/dt = Re. (2.11)
Expression (2.11) is a theorem on the change in the momentum of a mechanical system in differential form (in vector expression): the time derivative of the momentum vector of a mechanical system is equal to the main vector of all external forces acting on the system.
Projecting the vector equality (2.11) onto the Cartesian coordinate axes, we obtain expressions for the theorem on the change in the momentum of a mechanical system in a coordinate (scalar) expression:
dQ x /dt = R x e;
dQ y /dt = R y e;
dQ z /dt = R z e, (2.12)
those. the time derivative of the projection of the momentum of a mechanical system onto any axis is equal to the projection onto this axis of the main vector of all external forces acting on this mechanical system.
Multiplying both sides of equality (2.12) by dt, we obtain the theorem in another differential form:
dQ = R e ×dt = δS e, (2.13)
those. the differential of momentum of a mechanical system is equal to the elementary impulse of the main vector (the sum of elementary impulses) of all external forces acting on the system.
Integrating equality (2.13) within the time range from 0 to t, we obtain a theorem on the change in the momentum of a mechanical system in a finite (integral) form (in vector expression):
Q - Q 0 \u003d S e,
those. the change in the amount of motion of a mechanical system over a finite period of time is equal to the total impulse of the main vector (the sum of the total impulses) of all external forces acting on the system over the same period of time.
Projecting the vector equality (2.14) onto the Cartesian coordinate axes, we obtain expressions for the theorem in projections (in a scalar expression):
those. the change in the projection of the momentum of the mechanical system on any axis over a finite period of time is equal to the projection on the same axis of the total impulse of the main vector (the sum of the total impulses) of all external forces acting on the mechanical system for the same period of time.
From the considered theorem (2.11) - (2.15) follow the following corollaries:
- If R e = ∑ F j e = 0, then Q = const– we have the law of conservation of the momentum vector of the mechanical system: if the main vector Re of all external forces acting on a mechanical system is equal to zero, then the momentum vector of this system remains constant in magnitude and direction and equal to its initial value Q0, i.e. Q = Q0.
- If R x e = ∑X j e =0 (R e ≠ 0), then Q x = const- we have the law of conservation of the projection onto the axis of the momentum of the mechanical system: if the projection of the main vector of all forces acting on the mechanical system on any axis is zero, then the projection onto the same axis of the momentum vector of this system will be a constant value and equal to the projection onto this axis initial momentum vector, i.e. Qx = Q0x.
The differential form of the theorem on the change in momentum of a material system has important and interesting applications in mechanics. continuum. From (2.11) one can obtain Euler's theorem.
The amount of motion of the system, as a vector quantity, is determined by formulas (4.12) and (4.13).
Theorem. The time derivative of the amount of motion of the system is equal to the geometric sum of all external forces acting on it.
In the projections of the Cartesian axes, we obtain scalar equations.
You can write a vector
(4.28)
and scalar equations
Which express the theorem on the change in the momentum of the system in integral form: the change in the momentum of the system over a certain period of time is equal to the sum of the impulses for the same period of time. When solving problems, equations (4.27) are more often used
Law of conservation of momentum
Theorem on the change of the kinetic moment
The theorem on the change in the angular momentum of a point relative to the center: the time derivative of the angular momentum of a point relative to a fixed center is equal to the vector moment acting on the point of force relative to the same center.
or 
(4.30)
Comparing (4.23) and (4.30), we see that the moments of the vectors and are connected by the same dependence as the vectors themselves and are connected (Fig. 4.1). If we project equality onto the axis passing through the center O, we get
(4.31)
This equality expresses the theorem of the moment of momentum of a point about an axis.
|
(4.32)
If we project the expression (4.32) onto the axis passing through the center O, then we obtain an equality that characterizes the theorem on the change in the angular momentum relative to the axis.
(4.33)
Substituting (4.10) into equality (4.33) we can write differential equation rotating solid body (wheels, axles, shafts, rotors, etc.) in three forms.
(4.34)
(4.35)
(4.36)
Thus, it is advisable to use the theorem on the change in the kinetic moment to study the motion of a rigid body, which is very common in technology, its rotation around a fixed axis.
The law of conservation of the angular momentum of the system
1. Let in expression (4.32) .
Then it follows from equation (4.32) that , i.e. if the sum of the moments of all external forces applied to the system relative to a given center is zero, then the kinetic moment of the system relative to this center will be numerically and in direction will be constant.
2. If , then . Thus, if the sum of the moments of external forces acting on the system with respect to some axis is equal to zero, then the kinetic moment of the system with respect to this axis will be a constant value.
These results express the law of conservation of angular momentum.
In the case of a rotating rigid body, it follows from equality (4.34) that if , then . From here we come to the following conclusions:
If the system is immutable (absolutely solid), then , therefore, and and the rigid body rotates around a fixed axis with a constant angular velocity.
If the system is changeable, then . With an increase (then the individual elements of the system move away from the axis of rotation), the angular velocity decreases, because , and increases with decreasing, thus, in the case of a variable system, with the help of internal forces, it is possible to change the angular velocity.
Second task D2 control work is devoted to the theorem on the change in the kinetic moment of the system about the axis.
Task D2
A homogeneous horizontal platform (round with radius R or rectangular with sides R and 2R, where R = 1.2 m) with mass kg rotates with an angular velocity around the vertical axis z, which is spaced from the center of mass C of the platform at a distance OC = b (Fig. D2.0 – D2.9, table D2); dimensions for all rectangular platforms are shown in fig. D2.0a (top view).
At the moment of time, a load D with a mass of kg begins to move along the platform chute (under the action of internal forces) according to the law , where s is expressed in meters, t is in seconds. At the same time, a pair of forces with a moment M (given in newtonometers) begin to act on the platforms; at M< 0 его направление противоположно показанному на рисунках).
Determine, neglecting the mass of the shaft, the dependence i.e. platform angular velocity as a function of time.
In all figures, the load D is shown in a position where s > 0 (when s< 0, груз находится по другую сторону от точки А). Изображая чертеж решаемой задачи, провести ось z на заданном расстоянии OC = b от центра C.
Directions. Task D2 - on the application of the theorem on the change in the angular momentum of the system. When applying the theorem to a system consisting of a platform and a load, the angular momentum of the system about the z-axis is defined as the sum of the moments of the platform and the load. In this case, it should be taken into account that the absolute speed of the load is the sum of the relative and portable speeds, i.e. . Therefore, the amount of movement of this cargo 
. Then we can use the Varignon theorem (statics), according to which ; these moments are calculated in the same way as the moments of forces. The course of the solution is explained in more detail in example D2.
When solving the problem, it is useful to depict on the auxiliary drawing a view of the platform from above (from the end z), as is done in Fig. D2.0,a - D2.9,a.
The moment of inertia of a plate with mass m about the axis Cz, perpendicular to the plate and passing through its center of mass, is equal to: for a rectangular plate with sides and
;
For a round insert of radius R
| Condition number | b | s = F(t) | M |
| R R/2 R R/2 R R/2 R R/2 R R/2 | -0.4 0.6 0.8 10t 0.4 -0.5t -0.6t 0.8t 0.4 0.5 | 4t -6 -8t -9 6 -10 12 |
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Example D2. A homogeneous horizontal platform (rectangular with sides 2l and l), having a mass, is rigidly fastened to a vertical shaft and rotates with it around an axis z with angular velocity (Fig. E2a ). At the moment of time, a torque M begins to act on the shaft, directed oppositely ; at the same time cargo D mass located in the gutter AB at the point WITH, starts to move along the chute (under the action of internal forces) according to the law s = CD = F(t).
Given: m 1 \u003d 16 kg, t 2= 10 kg, l\u003d 0.5 m, \u003d 2, s \u003d 0.4t 2 (s - in meters, t - in seconds), M= kt, where k=6 Nm/s. Determine: - the law of change of the angular velocity of the platform.
Solution. Consider a mechanical system consisting of a platform and a load D. To determine w, we apply the theorem on the change in the angular momentum of the system about the axis z:
(1)
Let us depict the external forces acting on the system: the forces of gravity of the reaction and the torque M. Since the forces and are parallel to the z axis, and the reactions intersect this axis, their moments relative to the z axis are equal to zero. Then, considering the positive direction for the moment (i.e., counterclockwise), we obtain 
and equation (1) will take this form.
