The system discussed in the theorem can be any mechanical system consisting of any bodies.
Statement of the theorem
The amount of motion (impulse) of a mechanical system is a quantity equal to the sum of the amounts of motion (impulses) 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 momentum of a 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 a system
If the sum of all external forces acting on the system is zero, then the amount of motion (momentum) of the system is a constant quantity.
,
we obtain the expression of the theorem on the change in the momentum of the system in differential form:
Having integrated both sides of the resulting equality over an arbitrarily taken period of time 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 external forces acting on the system is equal to zero.
(moment of momentum m 2 kg s −1)
Theorem on the change in angular momentum 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 force acting on the point relative to the same center.
dk 0 /dt = M 0 (F ) .
Theorem on the change in angular momentum relative to an axis
the time derivative of the moment of momentum (kinetic moment) of a material point relative to any fixed axis is equal to the moment of the force acting on this point relative 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 mass m , moving under the influence of force F (Figure 3.1). Let's write down and construct the vector of angular momentum (kinetic momentum) M 0 material point relative to the center O :
Let us differentiate the expression for the angular momentum (kinetic moment k 0) by time:
Because dr /dt = V , then the vector product V ⊗ m ⋅ V (collinear vectors V And m ⋅ V ) is equal to zero. In the same time d(m ⋅ V) /dt = F according to the theorem on the momentum of a 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 a 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 about the change in angular momentum (angular 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 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 about the change in angular momentum (kinetic momentum) of a material point relative to the axis: the time derivative of the moment of momentum (kinetic moment) of a material point relative to any fixed axis is equal to the moment of the force acting on this point relative to the same axis.
Let us consider the consequences following from Theorems (3.4) and (3.5).
Corollary 1. Consider the case when the force F during the entire movement of the point passes through the stationary center O (case of central force), i.e. When M 0 (F ) = 0. Then from Theorem (3.4) it follows that k 0 = const ,
those. in the case of a central force, the angular 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 a moving point is a flat curve, the plane of which passes through the center of this force.
Corollary 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 force acting on a point relative to any fixed axis is always zero, then the angular momentum (kinetic moment) of the point relative to this axis remains constant.
Proof of the theorem on the change in momentum
Let the system consist of material points with masses and accelerations. We divide all forces acting on the bodies of the system into two types:
External forces are forces acting from bodies not included in the system under consideration. The resultant of external forces acting on a material point with number i let's denote
Internal forces are the forces with which the bodies of the system itself interact with each other. The force with which on the point with the number i the point with the number is valid k, we will denote , and the force of influence i th point on k th point - . Obviously, when , then
Using the introduced notation, we write Newton’s second law for each of the material points under consideration in the form
Considering that 
and summing up all the equations of Newton’s second law, we get:
The expression represents the sum of all internal forces acting in the system. According to Newton’s third law, in this sum, each force corresponds to a force such that, therefore, it holds 
Since the entire sum consists of such pairs, the sum itself is zero. Thus, we can write
Using the notation for the momentum of the system, we obtain
By 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 equations obtained allows us to state: a change in the momentum of the system occurs only as a result of the action of external forces, and internal forces cannot have any influence on this value.
Having integrated both sides 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:
where and are the values of the amount of motion of the system at moments of time and, respectively, and is the impulse of external forces over a period of time. In accordance with what was said earlier and the introduced notations,
Since the mass of the point is constant, and its acceleration, equation (2), expressing the basic law of dynamics, can be represented in the form
Equation (32) simultaneously expresses the theorem about the change in the momentum of a point in differential form: the time derivative of the momentum of a point is equal to the sum of the forces acting on the point
Let a moving point have a speed at the moment of time and a speed at the moment. Then we multiply both sides of equality (32) by and take definite integrals from them. In this case, on the right, where the integration occurs over time, the limits of the integral will be and on the left, where the speed is integrated, the limits of the integral will be the corresponding speed values
Since the integral of is equal, the result is
The integrals on the right, as follows from formula (30), represent the impulses of the acting forces. Therefore it will finally be
Equation (33) expresses the theorem about the change in the momentum of a point in final form: the change in the momentum of a point over a certain period of time is equal to the sum of the impulses of all forces acting on the point over the same period of time.
When solving problems, instead of the vector equation (33), equations in projections are often used. Projecting both sides of equality (33) onto the coordinate axes, we obtain
In the case of rectilinear motion occurring along the axis, the theorem is expressed by the first of these equations.
Problem solving. Equations (33) or (34) allow, knowing how the speed of a point changes when a point moves, to determine the impulse of the acting forces (the first problem of dynamics) or, knowing the impulses of the acting forces, to determine how the speed of a point changes when moving (the second problem of dynamics). When solving the second problem, when forces are given, it is necessary to calculate their impulses. As can be seen from equalities (30) or (31), this can be done only when the forces are constant or depend only on time.
Thus, equations (33), (34) can be directly used to solve the second problem of dynamics, when the data and required quantities in the problem include: acting forces, time of motion of the point and its initial and final velocities (i.e. quantities) , and the forces must be constant or dependent only on time.
Problem 95. A point with a mass of kg moves in a circle with a numerically constant speed. Determine the impulse of the force acting on the point during the time during which the point passes a quarter of the circle
Solution. According to the theorem on the change in momentum, constructing geometrically the difference between these quantities of motion (Fig. 222), we find from the resulting right triangle
But according to the conditions of the problem, therefore,
For analytical calculation, using the first two of equations (34), we can find
Problem 96. A load that has a mass and lies on a horizontal plane is given (by a push) an initial speed. The subsequent movement of the load is slowed down by a constant force F. Determine how long it will take for the load to stop,
Solution. According to the problem data, it is clear that to determine the time of movement, you can use the proven theorem. We depict the load in an arbitrary position (Fig. 223). It is acted upon by the force of gravity P, the reaction of the plane N and the braking force F. By directing the axis in the direction of movement, we compose the first of equations (34)
In this case, the speed at the moment of stopping), a. Of the forces, only force F gives the projection onto the axis. Since it is constant, where is the braking time. Substituting all this data into equation (a), we get the required time
Differential equation of motion of a material point under the influence of force F can be represented in the following vector form:
Since the mass of a point m is accepted as constant, then it can be entered under the derivative sign. Then
Formula (1) expresses the theorem on the change in the momentum of a point in differential form: the first derivative with respect to time of the momentum of a point is equal to the force acting on the point.
In projections onto coordinate axes (1) can be represented as
If both sides (1) are multiplied by dt, then we get another form of the same theorem - the momentum theorem in differential form:
those. the differential of the momentum of a point is equal to the elementary impulse of the force acting on the point.
Projecting both parts of (2) onto the coordinate axes, we obtain
Integrating both parts of (2) from zero to t (Fig. 1), we have
where is the speed of the point at the moment t; - speed at t = 0;
S- impulse of force over time t.
An expression in the form (3) is often called the momentum theorem in finite (or integral) form: the change in the momentum of a point over any period of time is equal to the impulse of force over the same period of time.
In projections onto coordinate axes, this theorem can be represented in the following form:
For a material point, the theorem on the change in momentum in any of the forms is essentially no different from the differential equations of motion of a point.
Theorem on the change in momentum of a system
The quantity of motion of the system will be called the vector quantity Q, equal to the geometric sum (principal vector) of the quantities of motion of all points of the system.
Consider a system consisting of n material points. Let us compose differential equations of motion for this system and add them term by term. Then we get:
The last sum, due to the property of internal forces, is equal to zero. Besides,
Finally we find:
Equation (4) expresses the theorem on the change in momentum of the system in differential form: the time derivative of the system's momentum is equal to the geometric sum of all external forces acting on the system.
Let's find another expression for the theorem. Let in the moment t= 0 the amount of motion of the system is Q 0, and at the moment of time t 1 becomes equal Q 1. Then, multiplying both sides of equality (4) by dt and integrating, we get:
Or where:
(S- force impulse)
since the integrals on the right give impulses of external forces,
equation (5) expresses 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 of external forces acting on the system over the same period of time.
In projections on the coordinate axes we will have:
Law of conservation of momentum
From the theorem on the change in momentum of a system, the following important corollaries can be obtained:
1. Let the sum of all external forces acting on the system be equal to zero:
Then from equation (4) it follows that in this case Q = const.
Thus, if the sum of all external forces acting on the system is equal to zero, then the vector of the system’s momentum will be constant in magnitude and direction.
2. 01Let the external forces acting on the system be such that the sum of their projections onto some axis (for example Ox) is equal to zero:
Then from equations (4`) it follows that in this case Q = const.
Thus, if the sum of the projections of all acting external forces onto any axis is equal to zero, then the projection of the amount of motion of the system onto this axis is a constant value.
These results express law of conservation of momentum of a system. It follows from them that internal forces cannot change the total amount of motion of the system.
Let's look at some examples:
· Phenomenon about the return of the roll. If we consider the rifle and the bullet as one system, then the pressure of the powder gases during a shot will be an internal force. This force cannot change the total momentum of the system. But since the powder gases, acting on the bullet, impart to it a certain amount of motion directed forward, they must simultaneously impart to the rifle the same amount of motion in the opposite direction. This will cause the rifle to move backwards, i.e. the so-called return. A similar phenomenon occurs when firing a gun (rollback).
· Operation of the propeller (propeller). The propeller imparts movement to a certain mass of air (or water) along the axis of the propeller, throwing this mass back. If we consider the thrown mass and the aircraft (or ship) as one system, then the forces of interaction between the propeller and the environment, as internal ones, cannot change the total amount of motion of this system. Therefore, when a mass of air (water) is thrown back, the aircraft (or ship) receives a corresponding forward speed such that the total amount of motion of the system under consideration remains equal to zero, since it was zero before the movement began.
A similar effect is achieved by the action of oars or paddle wheels.
· R e c t i v e Propulsion. In a rocket (rocket), gaseous products of fuel combustion are ejected at high speed from the hole in the tail of the rocket (from the jet engine nozzle). The pressure forces acting in this case will be internal forces and they cannot change the total momentum of the rocket-powder gases system. But since the escaping gases have a certain amount of motion directed backwards, the rocket receives a corresponding forward speed.
Theorem of moments about an axis.
Consider the material point of mass m, moving under the influence of force F. Let us find for it the relationship between the moment of the vectors mV And F relative to some fixed Z axis.
m z (F) = xF - yF (7)
Similarly for the value m(mV), if taken out m will be out of brackets
m z (mV) = m(xV - yV)(7`)
Taking the derivatives with respect to time from both sides of this equality, we find
On the right side of the resulting expression, the first bracket is equal to 0, since dx/dt=V and dу/dt = V, the second bracket according to formula (7) is equal to
mz(F), since according to the basic law of dynamics:
Finally we will have (8)
The resulting equation expresses the theorem of moments about the axis: the time derivative of the moment of momentum of a point relative to any axis is equal to the moment of the acting force relative to the same axis. A similar theorem holds for moments about any center O.
Consisting of n material points. Let us select a certain point from this system Mj with mass m j. As is known, external and internal forces act on this point.
Let's apply it to the point Mj resultant of all internal forces F j i and the resultant of all external forces F j e(Figure 2.2). For a selected material point Mj(as for a free point) we write the theorem on the change in momentum in differential form (2.3):
Let us write similar equations for all points of the mechanical system (j=1,2,3,…,n).
Figure 2.2
Let's add it all up piece by piece 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 ∑m j ×V j =Q– the amount of motion of the mechanical system;
∑F j e = R e– the main vector of all external forces acting on the mechanical system;
∑F j i = R i =0– the main vector of 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 = R e. (2.11)
Expression (2.11) is a theorem about the change in momentum of a mechanical system in differential form (in vector expression): the time derivative of the vector of momentum 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 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 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 change from 0 to t, we obtain a theorem about the change in the momentum of a mechanical system in final (integral) form (in vector expression):
Q - Q 0 = S e,
those. the change in the momentum 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 during 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 a mechanical system onto any axis over a finite period of time is equal to the projection onto 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 during the same period of time.
The following corollaries follow from the considered theorem (2.11) – (2.15):
- If R e = ∑F j e = 0, That Q = const– we have the law of conservation of the vector of momentum of a mechanical system: if the main vector R e of all external forces acting on a mechanical system is equal to zero, then the vector of momentum of this system remains constant in magnitude and direction and equal to its initial value Q 0, i.e. Q = Q 0.
- If R x e = ∑X j e =0 (R e ≠ 0), That Q x = const– we have the law of conservation of the projection onto the axis of the momentum of a mechanical system: if the projection of the main vector of all forces acting on a mechanical system onto any axis is zero, then the projection onto the same axis of the vector of the momentum of this system will be a constant value and equal to the projection onto this axis initial vector of momentum, i.e. Q x = Q 0x.
The differential form of the theorem on the change in momentum of a material system has important and interesting applications in continuum mechanics. From (2.11) we can obtain Euler’s theorem.
Let a material point move under the influence of force F. It is required to determine the movement of this point relative to the moving system Oxyz(see complex motion of a material point), which moves in a known way in relation to a stationary system O 1 x 1 y 1 z 1 .
Basic equation of dynamics in a stationary system
Let us write down the absolute acceleration of a point using the Coriolis theorem
Where a abs– absolute acceleration;
a rel– relative acceleration;
a lane– portable acceleration;
a core– Coriolis acceleration.
Let's rewrite (25) taking into account (26)
Let us introduce the notation 
- portable inertia force, 
- Coriolis inertial force. 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 transfer and Coriolis forces of inertia must be added to the forces acting on a point.
General theorems on the dynamics of a material point
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. Momentum of a material point– vector quantity equal to the product of the mass of a point and its velocity vector
.
(29)
2. Force impulse
Elementary impulse of force– vector quantity equal to the product of the force vector and an elementary time interval
(30).
Then full impulse
.
(31)
At F=const we get S=Ft.
The total impulse for a finite period of time can be calculated only in two cases, when the force acting on a 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 impulse (29) and momentum (30) allows us to establish a quantitative relationship between them.
Let us consider the movement of a material point M under the action of an arbitrary force F along an arbitrary trajectory.
ABOUT 
UD: 
.
(32)
We separate the 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 among the given and unknown quantities there are the mass of a point, its initial and final speed, forces and time of movement.
Theorem on the change in angular momentum of a material point
M 
moment of momentum of a material point relative to the center is equal to the product of the modulus of the momentum of the point and the shoulder, i.e. the shortest distance (perpendicular) from the center to the 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 a given mass m moves under the influence of force F.
,
,
,
(38)
.
(39)
Let's 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 zero, then the angular momentum of the point relative to this center retains its magnitude and direction.
If 
, That 
.
The theorem and conservation law are used in problems involving curvilinear motion, especially under the action of central forces.
