For a material point, the basic law of dynamics can be represented as
Multiplying both sides of this relation on the left vectorially by the radius vector (Fig. 3.9), we obtain
(3.32)
On the right side of this formula we have the moment of force relative to point O. We transform the left side by applying the formula for the derivative of a vector product
But 
as a vector product of parallel vectors. After this we get
(3.33)
The first derivative with respect to time of the moment of momentum of a point relative to any center is equal to the moment of force relative to the same center.
 |
An example of calculating the angular momentum of a system. Calculate the kinetic moment relative to point O of a system consisting of a cylindrical shaft of mass M = 20 kg and radius R = 0.5 m and a descending load of mass m = 60 kg (Figure 3.12). The shaft rotates around the Oz axis with an angular velocity ω = 10 s -1.
Figure 3.12
; ; 
For given input data, the angular momentum of the system
Theorem on the change in the angular momentum of a system. We apply the resultant external and internal forces to each point of the system. For each point of the system, you can apply the theorem on the change in angular momentum, for example in the form (3.33)
Summing over all points of the system and taking into account that the sum of derivatives is equal to the derivative of the sum, we obtain
By determining the kinetic moment of the system and the properties of external and internal forces
Therefore, the resulting relationship can be represented as
The first time derivative of the angular momentum of a system relative to any point is equal to the principal moment of external forces acting on the system relative to the same point.
3.3.5. Work of force
1) The elementary work of a force is equal to the scalar product of the force and the differential radius of the vector of the point of application of the force (Fig. 3.13)
Figure 3.13
Expression (3.36) can also be written in the following equivalent forms
where is the projection of the force onto the direction of the velocity of the point of application of the force.
2) Work of force on final displacement
Integrating the elementary work of force, we obtain the following expressions for the work of force on final displacement from point A to point B
3) Work of constant force
If the force is constant, then from (3.38) it follows
The work of a constant force does not depend on the shape of the trajectory, but depends only on the displacement vector of the point of application of the force.
4) Work of weight force
For the weight force (Fig. 3.14) and from (3.39) we obtain
Figure 3.14
If the movement occurs from point B to point A, then
In general
The “+” sign corresponds to the downward movement of the force application point, the “-” sign – upward.
4) Work of elastic force
Let the axis of the spring be directed along the x axis (Fig. 3.15), and the end of the spring moves from point 1 to point 2, then from (3.38) we obtain 
If the spring stiffness is With, so then
A 
(3.41)
If the end of the spring moves from point 0 to point 1, then in this expression we replace , , then the work of the elastic force will take the form
(3.42)
where is the elongation of the spring.
Figure 3.15
5) The work of force applied to a rotating body. The work of the moment.
In Fig. Figure 3.16 shows a rotating body to which an arbitrary force is applied. During rotation, the point of application of this force moves in a circle.
In some problems, instead of the momentum itself, its moment relative to some center or axis is considered as a dynamic characteristic of a moving point. These moments are defined in the same way as moments of force.
Momentum quantity of motion material point relative to some center O is called a vector defined by the equality
The angular momentum of a point is also called kinetic moment .
Momentum relative to any axis, passing through the center O, is equal to the projection of the momentum vector onto this axis.
If the momentum is given by its projections on the coordinate axes and the coordinates of the point in space are given, then the angular momentum relative to the origin is calculated as follows:
The projections of the angular momentum on the coordinate axes are equal to:
The SI unit of momentum is – .
End of work -
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Dynamics
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All topics in this section:
Unit systems
SGS Si Technical [L] cm m m [M]
Differential equations of motion of a point
The basic equation of dynamics can be written as follows
Basic tasks of dynamics
First or direct problem: The mass of a point and the law of its motion are known; it is necessary to find the force acting on the point. m
Most important cases
1. Force is constant.
Amount of point movement
The quantity of motion of a material point is a vector equal to the product m
Elementary and full force impulse
The action of a force on a material point over time
Theorem on the change in momentum of a point
Theorem. The time derivative of the momentum of a point is equal to the force acting on the point. Let's write down the basic law of dynamics
Theorem on the change in angular momentum of a point
Theorem. The time derivative of the moment of momentum of a point taken relative to some center is equal to the moment of force acting on the point relative to the same
Work of force. Power
One of the main characteristics of force that evaluates the effect of force on a body during some movement.
Theorem on the change in kinetic energy of a point
Theorem. The differential of the kinetic energy of a point is equal to the elementary work of the force acting on the point.
D'Alembert's principle for a material point
The equation of motion of a material point relative to an inertial reference system under the action of applied active forces and coupling reaction forces has the form:
Dynamics of a non-free material point
A non-free material point is a point whose freedom of movement is limited. Bodies that limit the freedom of movement of a point are called connections
Relative motion of a material point
In many dynamics problems, the motion of a material point is considered relative to a reference frame moving relative to an inertial reference frame.
Special cases of relative motion
1. Relative motion by inertia If a material point moves relative to a moving reference frame rectilinearly and uniformly, then such motion is called relative
Geometry of masses
Consider a mechanical system that consists of a finite number of material points with masses
Moments of inertia
To characterize the distribution of masses in bodies when considering rotational movements, it is necessary to introduce the concepts of moments of inertia. Moment of inertia about a point
Moments of inertia of the simplest bodies
1. Uniform rod 2. Rectangular plate 3. Uniform round disk
System movement quantity
The quantity of motion of a system of material points is the vector sum of quantities
Theorem on the change in momentum of a system
This theorem comes in three different forms. Theorem. The time derivative of the momentum of the system is equal to the vector sum of all external forces acting on
Laws of conservation of momentum
1. If the main vector of all external forces of the system is zero (), then the amount of motion of the system is constant
Theorem on the motion of the center of mass
Theorem The center of mass of a system moves in the same way as a material point, the mass of which is equal to the mass of the entire system, if all external forces applied to the point act on the point.
Momentum of the system
The angular momentum of a system of material points relative to some
Moment of momentum of a rigid body relative to the axis of rotation during rotational motion of a rigid body
Let us calculate the angular momentum of a rigid body relative to the axis of rotation.
Theorem on the change in angular momentum of a system
Theorem. The time derivative of the moment of momentum of the system, taken relative to some center, is equal to the vector sum of the moments of external forces acting on
Laws of conservation of angular momentum
1. If the main moment of the external forces of the system relative to the point is equal to zero (
Kinetic energy of the system
The kinetic energy of a system is the sum of the kinetic energies of all points of the system.
Kinetic energy of a solid
1. Forward movement of the body. The kinetic energy of a rigid body during translational motion is calculated in the same way as for one point whose mass is equal to the mass of this body.
Theorem on the change in kinetic energy of a system
This theorem comes in two forms. Theorem. The differential of the kinetic energy of the system is equal to the sum of the elementary works of all external and internal forces acting on the system
First, let's consider the case of one material point. Let be the mass of the material point M, be its speed, and be the amount of motion.
Let us select a point O in the surrounding space and construct the moment of the vector relative to this point according to the same rules by which the moment of force is calculated in statics. We get the vector quantity
which is called the angular momentum of the material point relative to the center O (Fig. 31).
Let us construct a Cartesian rectangular coordinate system Oxyz with the origin at the center O and project the vector ko onto these axes. Its projections onto these axes, equal to the moments of the vector relative to the corresponding coordinate axes, are called the moments of momentum of the material point relative to the coordinate axes:
Let us now have a mechanical system consisting of N material points. In this case, the angular momentum can be determined for each point of the system:
The geometric sum of the angular momentum of all material points that make up the system is called the principal angular momentum or kinetic moment of the system.
The amount of motion of the system, as a vector quantity, is determined by formulas (4.12) and (4.13).
Theorem. The derivative of the system's momentum with respect to time is equal to the geometric sum of all external forces acting on it.
In projections of Cartesian axes we obtain scalar equations.
You can write a vector
(4.28)
and scalar equations
Which express the theorem about 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 over the same period of time. When solving problems, equations (4.27) are more often used
Law of conservation of momentum
Theorem on the change in angular momentum
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 of force acting on the point relative to the same center.
Or 
(4.30)
Comparing (4.23) and (4.30), we see that the moments of the vectors and are related by the same dependence as the vectors and themselves are related (Fig. 4.1). If we project equality onto the axis passing through the center O, we get
(4.31)
This equality expresses the angular momentum theorem of a point relative to an axis.
|
(4.32)
If we project expression (4.32) onto the axis passing through the center O, we obtain an equality characterizing the theorem on the change in angular momentum relative to the axis.
(4.33)
Substituting (4.10) into equality (4.33), we can write the differential equation of a rotating rigid 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 kinetic moment to study the motion of a rigid body, which is very common in technology, its rotation around a fixed axis.
Law of conservation of angular momentum of a system
1. Let in expression (4.32) .
Then from equation (4.32) it follows that, i.e. if the sum of the moments of all external forces applied to the system relative to a given center is equal to zero, then the kinetic moment of the system relative to this center will be numerically and directionally constant.
2. If , then . Thus, if the sum of the moments of external forces acting on the system relative to a certain axis is zero, then the kinetic moment of the system relative 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 rigid body), then, consequently, the rigid body rotates around a fixed axis with a constant angular velocity.
If the system is changeable, then . With an increase (then individual elements of the system move away from the axis of rotation), the angular velocity decreases, because , and when decreasing it increases, thus, in the case of a variable system, with the help of internal forces it is possible to change the angular velocity.
The second task D2 of the test is devoted to the theorem on the change in the angular momentum of the system relative to the axis.
Problem D2
A homogeneous horizontal platform (circular with radius R or rectangular with sides R and 2R, where R = 1.2 m) with a mass of kg rotates with angular velocity around the vertical axis z, spaced from the center of mass C of the platform at a distance OC = b (Fig. E2.0 – 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 influence of internal forces) according to the law, where s is expressed in meters, t - in seconds. At the same time, a pair of forces with a moment M (specified in newtonometers; at M< 0 его направление противоположно показанному на рисунках).
Determine, neglecting the mass of the shaft, the dependence i.e. angular velocity of the platform as a function of time.
In all figures, the load D is shown in a position in which 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. Problem D2 – to apply 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 relative to the z axis is determined as the sum of the moments of the platform and the load. 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 load 
. Then you can use Varignon’s theorem (statics), according to which ; these moments are calculated in the same way as moments of forces. The solution is explained in more detail in example D2.
When solving a problem, it is useful to depict in an auxiliary drawing a view of the platform from above (from the z end), as is done in Fig. D2.0, a – D2.9, a.
The moment of inertia of a plate with mass m relative to 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 plate 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 10 t 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 attached 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 ; simultaneously cargo D mass located in the trench AB at the point WITH, begins to move along the chute (under the influence of internal forces) according to the law s = CD = F(t).
Given: m 1 = 16 kg, t 2= 10 kg, l= 0.5 m, = 2, s = 0.4t 2 (s - in meters, t - in seconds), M= kt, Where k=6 Nm/s. Determine: - the law of change in 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 relative to the axis z:
(1)
Let us depict the external forces acting on the system: the gravitational force 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 direction positive for the moment (i.e., counterclockwise), we obtain 
and equation (1) will take this form.
The direction and magnitude of the moment of momentum is determined in exactly the same way as in the case of estimating the moment of force (section 1.2.2).
At the same time we define ( main) angular momentum as the vector sum of the moments of the number of movements of the points of the system under consideration. It also has a second name - kinetic moment :
Let us find the time derivative of expression (3.40), using the rules for differentiating the product of two functions, and also the fact that the derivative of a sum is equal to the sum of derivatives (i.e., the sign of the sum can be moved as a coefficient during differentiation):
.
Let us take into account the obvious kinematic equalities: 
. Then: 
. We use the average equation from formulas (3.26) 
, and also the fact that the vector product of two collinear vectors ( and ) is equal to zero, we obtain:
Applying the property of internal forces (3.36) to the 2nd term, we obtain an expression for the theorem on the change in the main moment of momentum of a mechanical system:
 .
| (3.42) |
The time derivative of the kinetic moment is equal to the sum of the moments of all external forces acting in the system.
This formulation is often called briefly: moment theorem .
It should be noted that the theorem of moments is formulated in a fixed frame of reference relative to a certain fixed center O. If a rigid body is considered as a mechanical system, then it is convenient to choose the center O on the axis of rotation of the body.
One important property of the moment theorem should be noted (we present it without derivation). The theorem of moments is also true in a translationally moving reference system if the center of mass (point C) of the body (mechanical system) is chosen as its center:
The formulation of the theorem in this case remains practically the same.
Corollary 1
Let the right side of expression (3.42) be equal to zero =0, - the system is isolated. Then from equation (3.42) it follows that .
For an isolated mechanical system, the vector of the kinetic moment of the system does not change either in direction or in magnitude over time.
Corollary 2
If the right side of any of the expressions (3.44) is equal to zero, for example, for the Oz axis: =0 (partially isolated system), then from equations (3.44) it follows: =const.
Consequently, if the sum of the moments of external forces relative to any axis is zero, then the axial kinetic moment of the system along this axis does not change over time.
The formulations given above in the corollaries are the expressions law of conservation of angular momentum in isolated systems .
Momentum of a rigid body
Let's consider a special case - the rotation of a rigid body around the Oz axis (Fig. 3.4).
Fig.3.4
A point on a body separated from the axis of rotation by a distance h k, rotates in a plane parallel to Oxy at a speed of . In accordance with the definition of the axial moment, we use expression (1.19), replacing the projection F XY force on this plane by the amount of motion of the point 
. Let us estimate the axial kinetic moment of the body:
According to the Pythagorean theorem 
, therefore (3.46) can be written as follows:
 | (3.47) |
Then expression (3.45) will take the form:
| (3.48) |
If we use the law of conservation of angular momentum for a partially isolated system (Corollary 2) in relation to a solid body (3.48), we obtain . In this case, you can consider two options:
QUESTIONS FOR SELF-CONTROL
1. How is the angular momentum of a rotating rigid body determined?
2. How does the axial moment of inertia differ from the axial kinetic moment?
3. How does the rotation speed of a rigid body change over time in the absence of external forces?
Axial moment of inertia of a rigid body
As we will see later, the axial moment of inertia of a body has the same significance for the rotational motion of a body as the mass of a body during its translational motion. This is one of the most important characteristics of the body, determining the inertia of the body during its rotation. As can be seen from definition (3.45), this is a positive scalar quantity, which depends on the masses of the points of the system, but to a greater extent on the distance of the points from the axis of rotation.
For continuous homogeneous bodies of simple shapes, the value of the axial moment of inertia, as in the case of estimating the position of the center of mass (3.8), is calculated by the integration method, using the mass of an elementary volume instead of a discrete mass dm=ρdV:
 | (3.49) |
For reference, we present the values of the moments of inertia for some simple bodies:
m and length l relative to the axis passing perpendicular to the rod through its middle (Fig. 3.5).
Fig.3.5
The moment of inertia of a thin homogeneous rod with a mass m and length l relative to the axis passing perpendicular to the rod through its end (Fig. 3.6).
Fig.3.6
The moment of inertia of a thin homogeneous ring of mass m and radius R relative to the axis passing through its center perpendicular to the plane of the ring (Fig. 3.7).
Fig.3.7
The moment of inertia of a thin homogeneous disk with a mass m and radius R relative to the axis passing through its center perpendicular to the plane of the disk (Fig. 3.7).
Fig.3.8
· Moment of inertia of a body of arbitrary shape.
For bodies of arbitrary shape, the moment of inertia is written in the following form:
Where ρ - so-called radius of gyration body, or the radius of a certain conventional ring with mass m, the axial moment of inertia of which is equal to the moment of inertia of the given body.
Huygens–Steiner theorem
Fig.3.9
Let us associate two parallel coordinate systems with the body. The first Cx"y"z", with the origin at the center of mass, is called central, and the second Oxyz, with the center O, lying on the Cx" axis at a distance CO = d(Fig. 3.9). It is easy to establish connections between the coordinates of body points in these systems:
In accordance with formula (3.47), the moment of inertia of the body relative to the Oz axis:
Here the factors 2 are constant for all terms of the 2nd and 3rd sums of the right side d And d taken out of the corresponding amounts. The sum of the masses in the third term is the body mass. The second sum, in accordance with (3.7), determines the coordinate of the center of mass C on the axis Cx" (), and the equality is obvious: . Taking into account that the 1st term, by definition, is the moment of inertia of the body relative to the central axis Cz" (or Z C ) 
, we obtain the formulation of the Huygens - Steiner theorem:
 | (3.50) |
The moment of inertia of a body relative to a certain axis is equal to the sum of the moment of inertia of the body relative to a parallel central axis and the product of the mass of the body by the square of the distance between these axes.
QUESTIONS FOR SELF-CONTROL
1. Give formulas for the axial moments of inertia of a rod, ring, disk.
2. Find the radius of gyration of a round solid cylinder relative to its central axis.
