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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).
Number of movement 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 forward movement 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.
(in 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 motion of the center of mass are two different forms of the same 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.
An example of the calculation of a spur gear
An example of the calculation of a spur gear. The choice of material, the calculation of allowable stresses, the calculation of contact and bending strength were carried out.
An example of solving the problem of beam bending
In the example, plots are built transverse forces and bending moments, a dangerous section was found and an I-beam was selected. In the problem, the construction of diagrams using differential dependencies was analyzed, comparative analysis different cross sections of the beam.
An example of solving the problem of shaft torsion
The task is to test the strength of a steel shaft for a given diameter, material and allowable stresses. During the solution, diagrams of torques, shear stresses and twist angles are built. Self weight of the shaft is not taken into account
An example of solving the problem of tension-compression of a rod
The task is to test the strength of a steel rod at given allowable stresses. During the solution, plots of longitudinal forces, normal stresses and displacements are built. Self weight of the bar is not taken into account
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
Determination of the speed and acceleration of a point according to the given equations of motion
An example of solving the problem of determining the speed and acceleration of a point according to the given equations of motion
Determination of velocities and accelerations of points of a rigid body during plane-parallel motion
An example of solving the problem of determining the velocities and accelerations of points solid body in plane-parallel motion
Determination of Forces in Planar Truss Bars
An example of solving the problem of determining the forces in the bars of a flat truss by the Ritter method and the knot cutting method
Application of the torque change theorem
An example of solving the problem of applying the change theorem angular momentum to determine the angular velocity of a body rotating around a fixed axis.
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 amount of motion (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 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 the 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 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 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 the force acting on the point relative to any fixed axis is always equal to zero, then the angular momentum (kinetic moment) of the point relative to 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 acting 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
We write the absolute acceleration of a point according to the 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 of the momentum modulus of the point and the arm, 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 a 
, then 
.
The theorem and the conservation law are used in curvilinear motion problems, especially under the action of central forces.
Consider a system consisting of material points. Compose for this system differential equations motion (13) and add them term by term. Then we get
The last sum by the property of internal forces is equal to zero. Besides,
Finally we find
Equation (20) expresses the theorem on the change in the momentum of the 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 projections onto the coordinate axes it will be:
Let us find another expression of the theorem. Let at the moment of time the momentum of the system is equal to and at the moment it becomes equal to . Then, multiplying both sides of equality (20) by and integrating, we obtain
since the integrals on the right give the impulses of external forces.
Equation (21) 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 acting on the system of external forces over the same period of time.
In projections onto the coordinate axes it will be:
Let us point out the connection between the proved theorem and the theorem on the motion of the center of mass. Since , then, substituting this value into equality (20) and taking into account that we get , i.e. equation (16).
Therefore, the theorem on the motion of the center of mass and the theorem on the change in the momentum of the system are, in essence, two different forms the same theorem. In cases where the motion of a rigid body (or a system of bodies) is being studied, any of these forms can equally be used, and equation (16) is usually more convenient to use. For a continuous medium (liquid, gas), when solving problems, they usually use the theorem on the change in the momentum of the system. This theorem also has important applications in the theory of impact (see Ch. XXXI) and in the study of jet propulsion (see § 114).
Differential equation of motion of a material point under the action of a force F can be represented in the following vector form:
Since the mass of a point m is assumed to be constant, then it can be introduced under the sign of the derivative. Then
Formula (1) expresses the theorem on the change in the momentum of a point in differential form: the first time derivative of the momentum of a point is equal to the force acting on the point.
In projections onto the coordinate axes (1) can be represented as
If both sides of (1) are multiplied by dt, then we obtain 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- momentum of force over time t.
The 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 momentum of the force over the same period of time.
In projections onto the 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, in essence, does not differ from the differential equations of motion of a point.
Theorem on the change in the momentum of the system
The amount of motion of the system will be called the vector quantity Q, equal to the geometric sum (principal vector) of the momentum 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 by the property of internal forces is equal to zero. Besides,
We finally find:
Equation (4) expresses the theorem on the change in the momentum of the 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.
Let's find another expression of the theorem. Let at the moment t= 0 the momentum of the system is Q0, and at the moment of time t1 becomes equal Q1. 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 the 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 amount of motion 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 the momentum of the system, the following important consequences can be obtained:
1. Let the sum of all external forces acting on the system be equal to zero:
Then it follows from Eq. (4) that, in this case, Q=const.
In this way, 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 modulus and direction.
2. Let the external forces acting on the system be such that the sum of their projections on some axis (for example, Ox) is equal to zero:
Then it follows from equations (4`) that in this case Q = const.
In this way, if the sum of the projections of all acting external forces on some axis is equal to zero, then the projection of the momentum of the system on this axis is a constant value.
These results express law of conservation of momentum of the system. It follows from them that internal forces cannot change the total momentum of the system.
Let's look at some examples:
· P h e n i e o f recoil or recoil. If we consider a rifle and a bullet as one system, then the pressure of the powder gases when fired will be an internal force. This force cannot change the total momentum of the system. But since the powder gases, acting on the bullet, give it a certain amount of movement directed forward, they must simultaneously tell the rifle the same amount of movement in reverse direction. This will cause the rifle to move backward, i.e. so-called return. A similar phenomenon occurs when firing from a gun (rollback).
· Operation of the propeller (propeller). The propeller informs a certain mass of air (or water) of motion along the axis of the propeller, throwing this mass back. If we consider the ejected mass and the aircraft (or ship) as one system, then the interaction forces of the propeller and the medium as internal cannot change the total momentum of this system. Therefore, when a mass of air (water) is thrown back, the aircraft (or vessel) obtains the corresponding forward speed, such that the total momentum of the system under consideration remains equal to zero, since it was zero before the start of movement.
A similar effect is achieved by the action of oars or paddle wheels.
· Jet propulsion. In a rocket projectile (rocket), gaseous products of fuel combustion are ejected at high speed from a hole in the tail of the rocket (from the nozzle of a jet engine). 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 in this case the corresponding forward speed.
Theorem of moments about the axis.
Consider a material point of mass m moving under the influence of a force F. Let us find for it the dependence between the moment of the vectors mV and F about some fixed Z-axis.
m z (F) = xF - yF (7)
Similarly for the quantity m (mV), if taken out m bracket will be
m z (mV) \u003d m (xV - yV)(7`)
Taking time derivatives of both sides of this equality, we find
On the right side of the resulting expression, the first parenthesis is 0, since dx/dt=V and dу/dt=V, while the second bracket according to formula (7) is equal to
m z (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 angular momentum of a point about some axis is equal to the moment operating force about the same axis. A similar theorem also holds for moments about any center O.
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