The main theorems of the speakers Theoretical mechanics. System dynamics. Kinetic energy theorem
The use of OZMS in solving tasks is associated with certain difficulties. Therefore, additional relationships are used between the characteristics of motion and forces, which are more convenient for practical application. These ratios are general speaker theorems. They, as the consequences of OZMS, establish the relationship between the speed of the change of some specially entered movement measures and the characteristics of the external forces.
Theorem on the change in the amount of movement. We introduce the concept of the number of motion vector (R. Decart) of the material point (Fig. 3.4):
I І \u003d T V G. (3.9)
Fig. 3.4.
For the system we introduce the concept the main vector of the system movement As a geometric amount:
Q \u003d y, m "v r
In accordance with OZMS: Hyu, - ^ \u003d I), or X
R (E).
Taking into account, that / w, \u003d const We will get: -ym ,! "\u003d R (E),
or in final form
up to / d_ \u003d a (e (3.11)
those. the first time derivative of the main system of the system movement is equal to the main vector of external forces.
Theorem on the movement of the center of mass. Center Mass System refer to the geometric point whose position depends on t, and i.e. From the mass distribution / g /, in the system and is determined by the expression of the radius-vector of the center of the mass (Fig. 3.5):
where g s - Radius-vector center mass.
Fig. 3.5.
Let's call \u003d t with a mass system. After multiplying
(3.12) on denominator and differentiation of both parts
valuable equality will have: g s t with = ^ T.U. \u003d. 0, or 0 \u003d t with y with.
Thus, the main vector of the amount of system movement is equal to the mass of the mass system and the velocity of the mass center. Using the theorem about changing the amount of movement (3.11), we get:
t C reduction C / DI \u003d A (E), or
Formula (3.13) expresses the theorem about the movement of the center of mass: the center of the mass system is moving as a material point, which has a mass of the system, to which the main vector of external forces is operating.
Theorem on changing the moment of the amount of movement. We introduce the concept of the moment of the amount of material point movement as a vector product of its radius-vector and the amount of movement:
to oh, = bl H. t, U., (3.14)
where to OI - The moment of the magnitude of the material point relative to the fixed point ABOUT (Fig. 3.6).
Now we define the moment of the number of movement of the mechanical system as a geometric amount:
K () \u003d x k, \u003d Schu,? O-15\u003e
Differentiation (3.15), we get:
Ґ Сіk --- H. t і + g Y. H. t І.
Considering that \u003d Y y y і H. t і y \u003d 0, and formula (3.2), we get:
sIK A / C1ї - ї 0.
Based on the second expression in (3.6), we will finally have the theorem of changing the moment of the number of system movement:
The first time derivative of the time of the mechanical system relative to the fixed center is equal to the main point of the external forces acting on this system as to the same center.
When the relationship is derived (3.16) it was assumed that ABOUT - fixed point. However, it can be shown that in a number of other cases, the relationship (3.16) will not change, in particular, if with a flat movement a moment point to choose in the center of the masses, instantaneous speeds or accelerations. In addition, if the point ABOUT Coincides with a moving material point, equality (3.16), written for this point will appeal to the identity 0 \u003d 0.
The theorem on the change in kinetic energy. When moving the mechanical system, both "external" and the internal energy of the system is changed. If the characteristics of the internal forces, the main vector and the main moment, do not affect the change in the main vector and the main point of the amount of accelerations, then internal forces can be included in the estimates of the processes of the energy state of the system. Therefore, when considering changes in energy, the system has to consider the movements of individual points to which the internal forces are also attached.
The kinetic energy of the material point is determined as a magnitude
T ^ Tutug. (3.17)
The kinetic energy of the mechanical system is equal to the sum of the kinetic energies of material points of the system:
notice, that T\u003e 0.
We define power power as a scalar product of power vector on speed vector:
St. Petersburg State University civil aviation
Department No. 6 - "Mechanics"
Section III
"DYNAMICS"
St. Petersburg
- 2016 -1. Yablonsky A.A., Nikiforova V.M. Course
Theoretical mechanics. Static, kinematics,
dynamics. Textbook. M.: KNORS. 2011. - 608 p.
2. Meshchersky I.V. Tasks on theoretical
Mechanics. Studies. Benefit. St. Petersburg: Lan. 2011. - 448 p.
3. Targ M.S. Course of theoretical mechanics. M.:
High school. 2012. - 548 p.
4. Chernov K.I. Fundamentals of technical mechanics. M.:
Mechanical engineering. 1986. - 256 p.
5. ARET VA "Remote teaching
technology". (Electronic manual www.openmechanics.com), 2016.Galia 1. Introduction
in the dynamics. Laws and axioms
Dynamics of material point. Basic equation
Dynamics. Differential and natural equations
Movement. Two basic tasks of speakers. Examples
Solutions of the direct task of dynamics.
Lecture 2. Solution of the inverse dynamics problem. General
Note to solve the inverse speaker problem. Examples
Solutions of the inverse dynamics problem. Body movement
abandoned at an angle to the horizon, excluding resistance
air.
Lecture 3. Straight fluctuations in the material point.
Condition
occurrence
oscillations.
Classification
oscillations. Free oscillations excluding forces
Resistance.
Flowing
oscillations.
Decrement
oscillations.
Lecture 4. Forced vibrations of the material point.
Resonance.
Influence
Resistance
Movement
for
Forced oscillations. set 5.. Relative movement of material point.
Inertia forces. Private traffic cases for different
Views of a portable movement. Effect of Rotation of Earth on
Equilibrium and movement of tel.
Lecture 6. Dynamics of the mechanical system. Mechanical
system. External and domestic strength. Center mass system.
Theorem on the movement of the center of mass. Conservation laws.
An example of solving the problem of using the theorem
Movement center movement.
Lecture 7. Pulse force. The number of movement. Theorem Ob.
Changes in the amount of movement. Conservation laws.
Theorem Euler. An example of solving the task of using
Theorems on the change in the amount of movement. Moment
Motion. Moment Change theorem
quantities of movement ..
Lecture 8. Conservation laws. Elements of the theory of moments
inertia.
Kinetic
moment
hard
Body.
Differential solid rotation equation.
An example of solving the problem of using the theorem on
Change
Moment
number
Movement
Systems.
Elementary gyroscope theory.
Introduction to the dynamics
Lecture 1.Introduction to the dynamics
Dynamics - section of theoretical mechanics,
studying mechanical movement from the most common point
vision. Movement is regarded in connection with the existing
On the object forces.
The section consists of three departments:
Dynamics
Dynamics
Dynamics
material point
mechanical system
Analytical mechanics
Point Dynamics - Excelves Motion Motion
Taking into account the forces causing this movement.
Main object - material point - material
body having a mass of which can be
Neglect. The dynamics of the mechanical system - is studying the movement
the totality of material points and solids,
united by the general laws of interaction, taking into account
Forces causing this movement.
Analytical mechanics - studying the movement of non-free
mechanical Systems Using common
Analytical methods.
Main assumptions:
- There is an absolute space (it has purely
geometric properties that are independent of matter and
her movement);
- There is absolute time (independent of matter and
Her movements). From here implies:
- There is an absolutely fixed reference system;
- time does not depend on the movement of the reference system;
- masses of moving points do not depend on movement
reference system.
These assumptions are used in classical mechanics,
Created by Galileema and Newton. She has so far
A fairly wide range of applications, since
Consisting in Applied Sciences Mechanical
systems do not possess such large masses and
speed speeds for which their
influence on the geometry of space, time, movement, like
This is done in relativistic mechanics (theory
relativity). Strength - the value variable and depends on:
a) time - f f (t),
b) position of the point of the application of force - F F (R),
c) movement speed
Power application points - F f (V).
The material point may be free if
Moving does not impose restrictions. Otherwise,
The material point is called non-free
Inertness is the property of the material body faster or
slower change the speed of your movement
Under the action of the forces attached to him
Inertial reference systems are such systems,
where the law of inertia is performed; Otherwise, systems
The reference is non-intersocial
13. Basic Shaver
Gravity.F Mg.
G 9.81 m. / C2
acceleration of gravity
F f n Normal reaction.
friction coefficient
F 6.673 10-11 m3 / (kg C2).
F f m1m2 R 2
Slip friction force
Power of gravity.
gravitational constant
The power of elasticity
F C.
Extension (compression) Springs (m)
Spring stiffness coefficient (N / M).
The power of viscous friction. F V.
Body speed
Wednesday density
Slow motion
Resistance coefficient
1
F CX SV 2
2
Power hydrodynamic
area
The coefficient of resistance transverse
Resistance.
Sections
Fast traffic 14. Laws and axioms of the dynamics of the matt point
Based on classical mechanics lie laws
The "Mathematical began" in the work of I. Newton
Natural philosophy »(1687).
Basic laws of speakers - for the first time open by Galileem and
Newton formulated constitute the basis of all methods
Descriptions and analysis of the movement of mechanical systems and their
Dynamic interaction under the action of various forces.
Inertia Law (Galileo-Newton Law) - Isolated
The material point of the body retains its rest
or uniform straight movement until,
The attached forces will not make it change it.
Hence the equivalence of the state of rest and movement
By inertia (the law of relativity of Galilee). Reference system
in relation to which the law of inertia is performed,
called inertial. Property material point
strive to keep the continued speed of your movement
(its kinematic state) is called inertia. The law of proportionality of strength and acceleration
(The main equation of dynamics - II Law Newton) -
Acceleration reported by the material point by force
directly proportional to strength and back
in proportion to the mass of this point: A 1 F or MA
M.
F.
Here m - the mass of the point (measure of inertness), is measured in kg,
numerically equal to the weight divided by the acceleration of free
Falls:
G.
M.
G.
.
F - current force, measured in H (1N reports point
Weighing 1 kg acceleration 1 m / c2, 1 H \u003d 1 / 9.81 kgf). The law of equality of action and counteraction (III law
Newton) - any action corresponds to equal
the most oppositely directed
Countering:
M.
F2,1 M.
F1,2
F1, 2 F2,1
1
2
The law is fair for any kinematic state.
tel. Interaction forces, being attached to different
Points (bodies) are not balanced.
Law of Independence Action Forces - Acceleration
material point under the action of several forces
Equal to the geometric amount of point accelerations from
The actions of each of the forces separately:
A (F1, F2, ...) A1 (F1) A2 (F2) ....
or
A (R) A1 (F1) A2 (F2) .... 15. Basic dynamics equation
The main law of dynamics: the product of the mass of material
points to its acceleration that it gets under action
Forces, equal to the module of this force, and the direction of acceleration
coincides with the direction of the strength vector
MA F.
or
MA FK.
N.
The main dynamics equation: MA Fi (1).
- corresponds to the vector method of setting the movement of the point. 15.1. Differential traffic equations
material point
We substitute the acceleration of the point when typing
Movement
D 2r.
A.
dt.
2
.
2
D.
In the main equation of speakers: M R
Fi
2
dt.
(2) - Differential
point equation in
vector.
(2).
M.
F1.
F2.
R.
O.
A. In the coordinate form: we use the connection of the radius-vector with
Coordinates and vector of power with projections:
R (t) x (t) i y (t) j z (t) k
Fixi Fiy J FIZ K
d2.
After grouping
M 2 (Xi YJ ZK) (Fixi Fiy J FIZ K).
Vector relationship
dt.
Disintegrate
D 2x.
M x FIX;
Ox
:
M.
F.
;
IX.
2
on three scalar
dt.
M y fiy;
or
2
Equations:
D Y.
Z.
Oy.
:
M.
Fiy;
2
az.
M Z FIZ.
dt.
M (x, y, z)
R.
O.
I.
X.
K.
AY.
AX.
D 2Z.
(OZ): M 2 Fiz. - Differential
dt.
Motion equations
Z.
J.
X.
Y.
Y.
Points in coordinate
form.
This result can be obtained.
formal projection vector
Differential equation (1). Natural motion equations of the material point
- obtained by projection vector
differential equation of motion on natural
(movable) axis coordinates:
m s fi;
(): MAτ τ FIO;
(n): MAN FIN; or
S 2.
M.
FIN.
(b): M 0 FIB.
S.
O1 N.
F2.
- Natural
equations
Movement
Points.
B.
M.
A.
F1.
- Natural
Motion equations
Points. 16. Two basic tasks of the speakers
Direct Task: Movement (Motion Equations,
trajectory). It is required to determine the forces under the action
which occurs the specified movement.
Feedback: Forces are given, under the action of which
Moves move. Requires parameters
Y.
Movement
(equations of motion, motion trajectory).
Both tasks are solved using the basic dynamics equation and
Projection of it on coordinate axes. If the movement is considered
non-free point, as in the static, the principle is used
liberty from connections. As a result, the reaction of links turns on
The composition of the forces acting on the material point. The decision is first
Tasks are connected
Differentiation operations. Defense solution
R.
The tasks requires the integration of the corresponding differential
equations and it is much more complicated than differentiation.
The inverse task is more difficult to directly task Solving the direct task of the speakers - consider on
Examples:
Example 1. Elevator cabin weighing G rises with a cable with
Acceleration a. Determine the tension of the cable.
Solution: 1. Select the object (the elevator cabin moves progressively and
It can be viewed as a material point).
2. Return the connection (cable) and replace the reaction R.
3. We compile the main dynamics equation: MA FI G R
Y.
4. We project the main equation of speakers on the Y axis:
R.
(OY): May R g.
With a uniform movement of the cockpit AY \u003d 0 and the tension of the cable
Equally weight: T \u003d G.
A.
When cutting the cable T \u003d 0 and the acceleration of the cockpit is equal to acceleration
free fall: ay \u003d -g.
G.
AY.
G.
O.
R g Ma y g a y g (1).
Determine the reaction of the cable:
G.
G.
Determine the cable tension:
T R; T R G (1
AY.
G.
).The solution of the inverse dynamics problem is generally
Movement point of force acting on the point are
variables, depending on time, coordinates and speed.
The movement of the point is described by the three system
M x FIX;
second-order differential equations: m y f;
IY.
After integration
each of them will be X F1 (T, C1, C 2, C3); X F 4 (T, C1, C 2, ..., C 6); M Z FIZ.
six permanent Y F 2 (T, C1, C 2, C3); y f (t, c, c, ..., c); x x; y y; z z;
5
1
2
6
0
0
0
C1, C2, ...., C6:
z F 3 (T, C1, C 2, C3).
z F 6 (T, C1, C 2, ..., C 6). x x; y y; z z.
0
0
0
Values \u200b\u200bof constant C1, C2, ...., C6
are from six initial
x f1 (t, x 0, y 0, z 0); x f 4 (t, x 0, y 0, z 0, x0, y 0, z 0);
conditions at t \u003d 0:
After substitution of the found y f 2 (t, x 0, y 0, z 0); y f 5 (t, x 0, y 0, z 0, x0, y 0, z 0);
Permanent values \u200b\u200bGet: z f (t, x, y, z). z f 6 (t, x 0, y 0, z 0, x0, y 0, z 0).
3
0
0
0
Thus
way, under the action of the same strength system
X.
The material point can perform a whole class of movements,
Defined initial conditions.
The initial coordinates take into account the initial position of the point. Initial
The speed asked by projections, takes into account the effect on its movement
under consideration of the trajectory of the forces acting on the point before
coming to this area, i.e. Primary kinematic condition. 17. General instructions for solving direct and reverse
tasks. Order decision
1. Drafting the differential equation of movement:
1.1. Choose coordinate system - rectangular
(motionless) with an unknown trajectory of movement,
Natural (movable) with a well-known trajectory,
For example, a circle or straight line. In the latter case
You can use one straight line coordinate. Start
Coupling to combine with the initial position of the point (at t \u003d 0)
or with an equilibrium point position, if it exists,
For example, when swinging point. 1.2. Picture a point in the position corresponding to
arbitrary time (at T\u003e 0) so that
The coordinates were positive (s\u003e 0, x\u003e 0). Wherein
We also believe that the projection of speed in this position
also positive. In case of oscillations projection speed
changes the sign, for example, when returning to the Regulation
Equilibrium. Here it should be happened that in the considered
The point of time the point is removed from the equilibrium position.
The implementation of this recommendation is important in the future
Work with resistance forces depending on speed.
1.3. Release the material point from connections, replace
Their action by reactions, add active forces.
1.4. Record the basic law of speakers in vector form,
to proper on the selected axes, express the specified
or reactive forces at time time coordinates
Or speed if they depend on them. 2. Solution of differential equations:
2.1. Lower the derivative if the equation is not
It is provided to canonical (standard) mind.
eg:
DV X or S DV.
X.
,
dt.
dt.
2.2. Split variables, for example:
DVX
1
DVX
1
DV
K.
KDT or
G V 2,
KVX,
vx.
M.
dt.
M.
dt.
M.
DV
DT.
K 2.
G V.
M.
2.3. If there are three variables in equation,
That make the replacement of variables, for example:
DV X.
1
CX
dt.
M.
DV X DX V x DV X
1
CX.
DTDX
DX.
M.
And then split variables. 2.4. Calculate uncertain integrals in the left and
The right parts of the equation, for example:
DV X.
1
VX M KDT.
1
LN V x KT C1
M.
Using the initial conditions, for example, T \u003d 0, VX \u003d VX0,
Determine continuous integration:
1
ln v x V k t 0 c1; C1 LN V x 0.
x0.
M.
Comment. Instead of calculation uncertain integrals can
Calculate certain integrals with variable upper
limit
Lower limits represent the initial values \u200b\u200bof variables
(initial conditions). It is not required to separate
constant, which is automatically enabled in solution, for example:
V.
T.
DV
1
V.
M KDT.
v 0
0
LN V.
V.
v 0
1 T.
kt 0;
M.
ln v ln v 0
1
1
kt 0; Ln v kt ln v 0.
M.
M. 2.5. Express speed through the derivative coordinate
time, for example,
and repeat
1
KT LN V 0
ds.
Paragraphs 2.2 -2.4.
M.
V.
dt.
E.
Comment. If the equation is driven to canonical
form having a standard solution, then this is ready
The solution is used.
Permanent integration are still located
initial conditions. 18. Dynamics of the free material point
Movement of the point abandoned at an angle to the horizon in
uniform field of gravity without taking into account
Air resistance
DV X.
0;
Ox
:
M.
X.
0
;
dt.
MA.
F G.
I.
(OY): M y g Mg;
DV Y.
dt.
dv x 0; DV Y GDT;
vx.
vy
T.
VX 0.
VY0.
0
dv x 0; DV Y GDT;
V x V x0 V0 COS;
Y.
v0.
O.
X.
G.
X.
g;
DX.
V0 COS;
dt.
x v0 cos t;
V y v y 0 gt v0 sin gt;
DY.
V0 SIN GT;
dt.
GT 2.
y v0 sin t
;
219. Types of vibrations of the material point
1. Free oscillations (excluding resistance
media).
2. Free oscillations with the resistance of the medium
X.
(Flowing oscillations).
3. Forced oscillations.
4. Forced oscillations with resistance
medium.
Free oscillations - occur under action
Only restoring power.
We write the main law of dynamics: Ma g n r.
Select the coordinate system with the center in the position
Equilibrium (point O) and sprob
X axis equation:
O.
M x R cx.
L.
Y.
N.
R.
X.
X.
G.
We give the resulting equation
C.
To standard (canonical) mind: x k 2 x 0, where k 2.
M. This equation is homogeneous linear
Differential equation II order, view
whose solutions are determined by roots
The characteristic equation obtained by
Universal substitution: X E ZT.
X ZX2 E ZT.
z 2 k 2 0.
The roots of the characteristic equation
Imaginary and equal: z1, 2 ki.
General solution differential
The equations looks: X C1 COS KT C2 SIN KT.
Point speed: X KC SIN KT KC COS KT.
1
2
Starting conditions: T 0 x x0, x x 0.
Determine
Permanent: X0 C1 COS K 0 C2 SIN K 0 C11 C2 0.
X KC1 SIN K 0 KC2 COS K 0 KC1 0 KC21.
C1 x0.
C2.
x 0
.
K. Flowing oscillations of the material point -
The oscillatory movement of the material point occurs
in the presence of regenerating strength and power
Resistance to movement.
Dependence of the resistance force from displacement
or speed is determined by the physical nature of the medium or
Communication interfere with movement. The simplest
dependence is linear dependence on speed
(viscous resistance).
The attenuation of oscillations occurs very quickly. Basic
Influence of the forces of viscous resistance - reduction
The amplitudes of oscillations over time. 20. Relative motion of the material point
We put that the moving (non-intersocial) coordinate system Oxyz is moving
Some law relative to the fixed (inertial) coordinate system
O1x1y1z1. Motion of material point M (x, y, z) relatively mobile
OxyZ-relative systems, relatively fixed O1x1y1Z1 system
Absolute. The movement of the mobile system Oxyz is relatively fixed
Systems O1x1y1z1- Portable Movement.
Absolute
The main dynamics equation: MA Fi. Acceleration point:
M (A A A) Fi.
R.
E.
C.
A A A R A E A C.
We transfer the terms with portable and
R.
E.
C.
Coriolis acceleration to the right side: MA FI MA MA.
Transferred terms have the dimension of the forces and
are considered as appropriate forces
E MA E, C MA C.
Inertia, equal:
R.
In the projections on the axis of the mobile system
Ma Fi E C.
Coordinates we have:
F.
F.
(OZ): M Z \u200b\u200bF
Then the relative movement of the point
(OX): M x
can be considered as absolute
If you add to the current forces
(OY): M y
Portable and Coriolis forces inertia:
IX.
ex cx;
IY.
Ey CY;
IZ.
EZ CZ.
Thanks for attention!
Lecture 2.
21. The dynamics of the mechanical systemSystem of material dots or mechanical system -
A combination of material points or material bodies,
united by common interaction laws (position
or the movement of each of the points or body depends on the position
And the movements of everyone else).
System of free points - whose movement is not
limited to any connections (for example, planetary
The system in which planets are treated as
material dots).
Non-free dots or non-free
Mechanical system - Movement of material points or
bodies are limited to the system connected
(for example, mechanism, machine, etc.).
Lecture 2.
22. Forces acting on the systemIn addition to the previously existing classification of forces
(active and reactive forces) is introduced new
Classification of forces:
1. External forces (E) - acting on points and bodies
systems from dots or bodies that are not part
This system.
2. Internal forces (I) - interaction forces
material dots or bodies included in this
System.
The same force may be both external and
internal power. It all depends on which mechanical
The system is considered.
For example: in the system Sun, Earth and Moon all forces
Between them are internal. For
Consideration of the system Earth and the Moon of Power
Applied by the sun - external. Based on the law of action and counteracting each
The inner strength fk corresponds to another internal
FK force "equal to the module and the opposite
Direction.
Two remarkable properties of the internal forces are followed from this:
1. The main vector of all the internal forces of the system is equal
I.
I.
zero: R fk 0.
2. The main moment of all internal forces of the system
I.
I.
M.
M.
KO 0.
relative to any center is zero: o
BUT
IN
Z.
X ki 0; Yki 0; Z ki 0.
I.
I.
I.
M.
0
;
M.
0
;
M.
KX.
KY.
KZ 0.
FROM
Note: Although these equations are similar to equilibrium equations, they
these are not because the internal forces are applied to
various points or system bodies and can cause the movement of these
Points (tel) relative to each other. From these equations should
that internal forces do not affect the movement of the system considered
as one. 23. Mass Center of Materials System
To describe the movement of the system as a whole introduced
The geometric point, called the center of mass, the radius of which is determined by the expression
MK RK.
R.
,
C.
where m is the mass of the entire system:
M mk.
M.
Or in projections on coordinate axes:
MK XK.
XC.
,
MK Y K.
YC.
,
M.
z M1.
R1
RC
m2.
O.
X.
YC.
MK.
C R.
K.
zc.
R2.
M.
RN.
XS.
MN.
Y.
MK Z K.
zc.
.
M.
Mass Center Formulas
similar to the formulas for the center
severity. However, the concept of the center
masses more general because it is not
associated with the forces of grave or
Forces of gravity. 24. Theorem on the movement of the center of the mass system
Mk a k f k f k or mk
E.
I.
2
D.
E.
M.
R.
R.
.
2 K K.
dt.
In projections
Coordinate axes:
D 2 RK.
dt.
2
FKE FKI. Summing up
These equations
For all points:
Mrc mk rk.
d2.
E.
M.
R.
R.
.
C.
2
dt.
MK.
D 2 RK.
DT 2.
FKE FKI.
Re.
M.
D 2 RC
DT 2.
Re.
Ri 0.
Mac R.
M x C R EX FXKE; Theorems: Work
M y C R EY
M Z C R EZ
Mass system by
FYKE; Accelerating its center
Mass is the main thing
E.
FZK. Vector of external forces.
E. Consequences of the theorem on the movement of the center of the mass system
(conservation laws)
equal to zero, Re \u003d 0, then the speed of the center of mass is constant, VC \u003d const (center
Mass moving uniformly straightforward - the law of conservation of motion
center of mass).
2. If in the time interval, the projection of the main vector of external
System forces on the x axis equal to zero, Rxe \u003d 0, then the speed of the center of the masses along the x axis
equal to zero, Re \u003d 0, and at the initial moment the speed of the center of the masses is equal
zero, vc \u003d 0, then the radius-vector of the center of the masses remains constant, Rc \u003d
Const (the center of the masses is alone - the law of preserving the situation
center of mass).
The forces of the system on the x axis is zero, Rxe \u003d 0, and at the initial moment speed
The center of mass along this axis is zero, vcx \u003d 0, then the coordinate of the center of mass
X axis remains constant, XC \u003d const (the center of the masses is not moving along this
axis).
25. Power pulse
Measuring mechanical interaction characterizing
Transmission mechanical Movement by acting
to the point forces during this period of time:
S F (T 2 T1).
In projections
T.
T.
T.
Coordinate (OX): S X FX DT; (OY): S y FY DT; (OZ): S Z FZ DT.
T.
T.
T.
Axis:
2
2
2
1
1
1
T2.
In case of constant strength: S F DT
T1.
S X FX (T 2 T1);
S y fy (T 2 T1);
S Z FZ (T 2 T1);
The impulse is equally equal to the geometric
the sum of the pulses applied to the point forces per and that
same time interval: R F1 F2 ... Fn.
R DT F1DT F2 DT ... FN DT.
We integrate on T2.
T2.
T2.
T2.
This gap R DT F1DT F2 DT ... FN DT.
T1.
T1.
T1.
T1.
Time:
S S1 S 2 ... S n. 26. The number of traffic movement
equal to the product of the mass point on the vector of it
Speed: Q MV.
The number of motion of the material dot system -
The geometric amount of the matters of material
Points: Q1 Q2 ... Qn QK.
By definition of the center of mass:
Q.
M.
V.
Q qk mk vk mk
DRK.
D.
(MK Rk).
dt.
dt.
Mrc mk rk.
The vector of the number of system movement is equal
The mass of the entire system on the speed vector
center of mass system.
DRC
D.
Then: Q DT (MRC) M DT MVC.
In projections
Q MX C;
Coordinate axes: x
Q MVC.
Q y MX C;
Q y MX C. 26. Theorem on the change in the amount of movement
Systems
Consider the system N of material points. Applied K.
every point of force is divided into external and internal and
We will replace them with the appropriate equal FKE and FKI.
We write the main dynamics equation for each point:
MK A K F KE F Ki or MK DVK FKE FKI.
dt.
Summing up these
In the left part of the equation we will make
equations
Mass under the sign of the derivative
For all points:
and replace the amount of derivatives on
DVK.
E.
I.
M.
F.
F.
.
K.
K.
K.
Derivative amount: D (M V) R e.
dt.
K K.
dt.
From the definition
E.
I.
D.
Q.
E.
R.
0
R.
The amount of MK V K Q.
R.
Derivative vector of the number of time system
DT is equal to the main vector of external system forces.
System Movement:
DQX
In projections on the coordinate DQX R E F E; DQX R E F E;
R E FXKE.
xk.
xk.
dt.
dt.
dt.
Axis:
X.
X.
X. 26. Consequences of the number of quantity change theorem
System movements (conservation laws)
:
1. If in the time interval, the main vector of external
The forces of the system is zero, Re \u003d 0, then the number of quantity
Movement constant, Q \u003d Const - the law of conservation
The number of system movement.
2. If in the time interval Projection of the main vector
The external forces of the system on the X axis is zero, Rxe \u003d 0, then
Projection of the number of system movement on the X axis
constant, qx \u003d const.
Similar statements are valid for axes Y and Z.
DQ.
We proact to the axis: τ M1 G COS M2 G COS 0.
dt.
Separate
Q.
T.
variables
DQτ (M1 G COS M2 G COS) DT 0.
0
And integrate: Q0
Hence the law Qτ Qτ 0 0 or Qτ 0 Qτ.
Preservation: MV M V M V.
1 1
2 2
Right integral
Almost equal
zero, because time
Explosion T.<<1.
v2.
MV M1V1
v2.
m2. 27. The moment of the number of traffic or kinetic
moment of movement relative to some center
Mechanical motion measured by vector
Equal vector product of radius-vector
material point on the vector of its number of movement:
Q.
V.
Kinetic moment of system of material points
relative to some center - geometric
The sum of the moments of the number of movements of all
material points relative to the same center:
M.
KO.
R.
O.
K O R Q R Mv.
K x y (mv z) z (mv y);
K y z (mv x) x (mv z);
K z x (mv y) y (mv x).
Motion quantity vector derivative
Systems relative to a certain time center
equal to the main moment of the external forces of the system
relative to the same center.
KO K1O K2O ... KNO KIO RI MI VI.
In projections
KX.
On the axis:
K.
IX.
; K y kiy;
K z kiy. 28. The current pointing theorem
System Movement
Consider the system N of material points. Applied K.
every point of force is divided into external and internal and
We will replace them with the appropriate equal FKE and FKI.
We write the main dynamics equation for each point:
DVK.
E.
I.
E.
I.
M.
F.
F.
.
mk a k f k f k
K.
or
K.
K.
dt.
Multiply vector Each of the equals on the radius-vector
left:
DV
RK MK.
K.
dt.
Summing up these
Equations in all
Points:
RK FKE RK FKI.
DVK.
E.
I.
R.
M.
R.
F.
R.
F.
K.
k k k k.
K.
dt.
E.
Mo.
I.
Mo.
0Let's see if it is possible to make a sign of the derivative
Outside vector work:
DRK.
DVK.
D.
RK MK VK)
MK VK RK MK
dt.
dt.
dt.
VK MK VK 0 (SIN (VK, MK VK) 0)
DVK.
RK MK.
.
dt.
D.
E.
R.
M.
V.
M.
K.
K K.
O.
dt.
Thus, received:
Replace the amount of derivatives
For derivative sum: D
(Rk Mk V k) M OE.
dt.
The expression in brackets is the moment of the amount of movement
Systems. From here:
dk
O.
dt.
M OE. In projections for coordinate axes:
DK Y.
DK X.
DK Z.
E.
E.
MX;
M y;
M Ze.
dt.
dt.
dt.
Theorem: Moment Vector Derivative
the amount of system movement relatively
Some time center is equal to the main
The moment of external system forces relatively
of the same center.
dk
O.
dt.
M OE.
Theorem: Derivative Moment
System movements relative to some axis
in time is equal to the main moment of external
System forces regarding the same axis.
DK Y.
DK X.
DK Z.
E.
E.
MX;
M y;
M Ze.
dt.
dt.
dt. 29. The consequences of the point of change theorem
The amount of system movement (conservation laws)
1. If in the time interval vector of the main point
external system forces relative to some center
equal to zero, moe \u003d 0, then the moment of the moment
Movement of the system relative to the same center
constant, ko \u003d const - the law of maintaining moment
The number of system movement).
2. If in the time interval, the main moment of external
system forces relative to the x axis is zero, mxe \u003d 0, then
The moment of the amount of system movement relative to the x axis
constant, kx \u003d const.
Similar statements are valid for axes Y and Z. 30. Elements of the theory of inertia moments
With the rotational movement of the solid body of the inertia
(resistances change change) is the moment
inertia relative to the axis of rotation. Consider the mains
Concepts of definition and methods for calculating moments
inertia.
30.1. Moment of inertia of the material point
Regarding axis
2
2
2
I z Mh M (x Y)
Z.
H.
M.
Z.
R.
O.
H.
X.
X.
Y.
Y.
Moment inertia material
points relative to the axis equal
Mass point on
Square distance point to the axis.
In addition to the axial torque of the inertia of the solid
There are other types of inertia moments:
I xy xydm
- centrifugal moment of inertia
solid body. 30.2. The moment of inertia of the solid body relative to the axis
Z.
I z mk hk2 mk (xk2 yk2)
hk.
RK.
MK.
Z.
Y.
O.
yk.
X.
The moment of solid inertia
relative to the axis equal to the amount
Mass of each point
On the square of the distance of this point
to the axis.
When moving from discrete
Small mass to infinitely small
Mass point limit of such a sum
Determined by the integral:
xk.
I z h 2 dm (x 2 y 2) dm
- Axial moment of inertia
solid body.
I O R DM (X Y Z) DM
2
2
2
2
- Polar moment
Inertia of a solid body. 30.4. The moment of inertia of a homogeneous rod of permanent
Sections relative to axis
We highlight the elementary volume DV \u003d ADX at a distance x:
ZS.
Z.
Elementary
weight:
Dm Adx
L.
X.
X.
C.
DX.
L.
3 L.
L.
X.
I z x 2 dm x 2 adx a
3
0
0
0
L3 ML2.
A.
3
3
the location of the axis and set the integration limits (-l / 2,
L / 2). Here we will demonstrate the formula for the transition to
Parallel axes:
2
2
ML.
L.
I zc m.
3
2
I z i zc d m.
2
I ZC.
2
ML L.
ML2.
M.
.
3
12
2
230.5. The moment of inertia of a homogeneous solid cylinder
Regarding the axis of symmetry
We highlight the elementary volume: DV \u003d 2πrDRH (thin cylinder
Radius R.
Elementary weight:
DM 2 RDRH
R.
R.
I Z R DM R 2 2 RDRH
2
0
0
4 R.
R.
2 H.
4
0
R 4 MR 2
2 H.
4
2
MR 2.
IZ.
2
Since the height of the cylinders as a result is not included in
Formulas of the moments of inertia, then they remain
Fair for a thin solid disk and rim
Wheels (thin ring). 31. Kinetic moment of solid body
Δk zi Hi ΔMi VI Hi ΔMi z hi h Δmi.
2
Z I.
K z Δk zi z h Δmi z i z.
2
I.
Or passing
To infinitely small:
DK Z HDMV HDM z h z h dm.
2
K z dk z z h 2 dm z i z.
Kinetic moment of rotating
Body is equal to the work of the corner
Speed \u200b\u200bat the time of inertia
Regarding the axis of rotation.
Z.
Z.
HI
ΔMi.
VI
X.
Y. 32. Differential rotation equation
solid body relative to the axis
We write the theorem about the change in the kinetic moment
solid, rotating around the stationary axis:
DK Z.
M Ze.
dt.
The kinetic torque of the rotating solid is equal to:
Z.
Z.
Z.
MZ.
X.
K z z i z.
Moment of external forces relative to the axis
rotation is equal to the torque
(Reactions and gravity M M M
Z.
Z.
Rotate.
Moments do not create):
We substitute the kinetic moment and
Y.
torque in theorem
D (z i z)
M z M rotation.
dt.
I z m z M rotation. 33. Elementary gyroscope theory
Gyro - solid, rotating around the axis
material symmetry, one of whose points
Fixed.
Free gyroscope - enshrined in such a center of mass
remains fixed, and the axis of rotation passes through
center of mass and can take any position in
space, i.e. The axis of rotation changes its position
Like the axis of its own rotation of the body when
spherical movement.
Kc.
ω The main assumption of approximate (elementary)
Gyroscope theory - vector momentum vector
movement (kinetic moment) rotor is considered
directed along its own axis of rotation.
The main property of the free gyroscope is the axis of the rotor
retains the unchanged direction in space by
relation to an inertial (star) reference system
(is demonstrated by the foco pendulum that preserves unchanged
In relation to the stars, the plane of swing, 1852).
This implies the law of preserving the kinetic moment.
relative to the center of the mass of the rotor, provided
Disranged by friction in the axis bearings of the suspension
Rotor, external and inner frame:
DK C.
M CE 0;
dt.
K C Const. 34. Effect of force on the axis of a free gyroscope
In case of force applied to the rotor axis,
The moment of external forces on the center of the masses is not equal
zero:
dk
M E FH.
C.
dt.
M CE R F;
C.
Kinetic moment derivative
equal to the velocity of the end of this vector (cut theorem):
DK C.
Dr.
v k; (v).
dt.
dt.
vk.
Z.
M CE.
This means that the axis of the rotor will be
deviate
forces, and in the direction of the moment of the moment
of this force, i.e. will rotate non.
relative to the x axis (internal
suspension), and relative to the Y axis
(outer suspension).
F.
H.
vk.
Y.
FROM
M CE
X.
ω
Kc. With the termination of the force of the Rotor axis will remain
in the constant position corresponding to
The last time of the force of force, because
From this point on, the moment of external forces again
It becomes equal to zero.
In case of short-term strength (impact) axis
A gyroscope practically does not change its position.
Thus, the rapid rotation of the rotor reports
gyroscope ability to counteract random
The impacts seeking to change the position of the axis
rotation rotor, and with constant force action
retains the position of the plane perpendicular
Active strength, in which the axis of the rotor lies. These properties
Used in the operation of inertial navigation systems.
Thanks for attention!
Example: two people M1 and M2 are in a boatWeighing M3. In the initial moment of time a boat with people
was alone. Determine the movement of the boat if
The man of M2 has moved to the nose of the boat at a distance a.
1. Movement object
(boat with people):
x2
Y.
x1
2. Rewind links (water):
but
G3.
3. Replace the connection with the reaction:
4. Add active forces:
G1.
R.
G2.
X.
O.
We project on the x axis:
M x C 0.
XC const.
Mac R E G1 G2 G3 N
0 m1b m2 a.
B.
m2.
a.
M1.
x3.
x C Const 0.
MK XK 0 mk xk.
M2 A.
0 M1L M2 (L a) M3L
L.
M1 M2 M3.
in the opposite direction.
17
Lecture 6 (continued 6.2)
The theorem on the movement of the center of the mass of the system - Consider the system N of material points. Applied to each point of force separatedOn external and internal and replace them on the appropriate equal FKE and FKI. We write to each point the main equation
Dynamics:
or
D 2 RK.
E.
I.
D 2 RK.
Thind these equations
Mk a k f ke f ki
MK.
F.
F.
.
M.
K 2 FKE FKI.
K.
K.
For all points:
DT 2.
dt.
In the left part of the equation we will make the mass under the sign of the derivative
d2.
(m r) R e.
and replace the amount of derivatives on the derivative sum:
2 K K.
From the determination of the center of mass:
After the mass of the system
for the sign of the derivative we get
In projections for coordinate axes:
Mrc mk rk.
M.
D 2 RC
dt.
2
dt.
Substitute to the resulting equation:
R e or:
M x C R EX X KE;
M y c r ey yke;
Mac R E.
d2.
(MRC) R E.
2
dt.
Re.
Ri 0.
Production of the mass system to accelerate its center mass
Equally the main vector of external forces.
The center of the mass system moves as a material point of mass equal to mass
The entire system to which all external forces acting on the system are applied.
Example: Two people with M1 and M2 masses are in a boat mass M3.
At the initial moment of time, the boat with people was alone.
Determine the movement of the boat, if the man of M2 has moved to the nose
Consequences of the theorem on the movement of the center of the mass system
Boats a distance a.
Y.
(Conservation laws):
x2
but
1. If in the time interval, the main vector of external system forces
X.
1. Motion object (boat with people):
1
equal to zero, Re \u003d 0, then the speed of the center of mass is constant, VC \u003d const
2. Rewind links (water):
(The center of the masses is moving uniformly straightforward - the law of conservation
3.
We replace the connection with the reaction:
G1.
X.
Movement of the center of mass).
O.
G2.
2. If in the time interval of the projection of the main vector of external forces 4. Add active forces:
The systems on the x axis are zero, Rxe \u003d 0, then the speed of the center of the masses along the x axis
5. We write down the theorem about the center of the masses:
constant, VCX \u003d const (center of masses moving along the axis evenly).
G3.
R.
Mac R E G1 G2 G3 N
Similar statements are valid for axes Y and Z.
x3.
3. If in the time interval, the main vector of external system forces
We proact to the X: M x C 0 axis.
x C Const 0.
equal to zero, Re \u003d 0, and at the initial moment the speed of the center of mass is zero,
XC const.
VC \u003d 0, then the radius-vector of the center of mass remains constant, RC \u003d const (center
MK XK 0 mk xk.
The masses are located alone - the law of preserving the situation of the center of the masses).
We define for what distance it is necessary to cross the mass M1 mass,
M1 x1 m2 x2 m3 x3 m1 (x1 l) m2 (x2 l a) m3 (x3 l)
4. If in the time interval, the projection of the main vector of external
forces
To boat remains in place:
The systems on the x axis are zero, Rxe \u003d 0, and at the initial moment the speed of the center
M2 A.
0 M1L M2 (L a) M3L
M1 X1PO this
M2 x2osi
Mravna
M1 (x1v cxb \u003d) 0,
M.
(x2 a) Mortar
L.
mass
TO2 coordinate
masses along the x axis
3 x3 zero,
3 x3.
M1 M2 M3.
m2.
It remains constant, XC \u003d const (the center of the masses is not moving along this axis).
The boat will move to distance L
B.
A.
.
0
M.
B.
M.
A.
.
Similar
Fair for the axes of YMI Z.
1 approval
2
17
in the opposite direction.
M Z C R EZ Z KE.
1
Lecture 8 (continued 8.2)
4.The moment of inertia of a homogeneous rod of permanent
Sections relative to the axis:
Highlight elementary
ZS.
Volume DV \u003d ADX
Z.
L.
At a distance x:
5.
The moment of inertia of a homogeneous solid cylinder
Regarding the symmetry axis:
Highlight elementary
Volume dv \u003d 2πrdrh
(thin radius cylinder R):
Elementary
weight:
DM 2 RDRH
Z.
R.
X.
DX.
L.
Elementary
weight:
DM.
C.
X.
L.
3 L.
X.
I z x dm x adx a
3
0
0
2
2
0
ADX.
H.
R.
L3 ML2.
A.
3
3
Y.
To calculate the moment of inertia relative to the central
axis (passing through the center of gravity) is enough to change
The location of the axis and set the integration limits (-l / 2, L / 2).
Here we will demonstrate the formula for the transition to parallel
Axes:
2
2
I z i zc d 2 m.
2
I ZC.
6.
ML2 L.
ML2.
M.
.
3
12
2
0
0
4 R.
X.
R.
R.
2 H.
4
Dr.
0
R 4 MR 2
2 H.
4
2
Here used the formula for the volume of the cylinder V \u003d πr2h.
To calculate the moment of inertia of the hollow (thick) cylinder
It is enough to set the integration limits from R1 to R2 (R2\u003e R1):
ML.
L.
I zc m.
3
2
R4.
I z 2 H
4
R2
R1
2
2
R24 R14 M (R 2 R1)
2 H.
.
4
4
2
The moment of inertia of the thin cylinder relative to the axis since the height of the cylinders as a result is not included in the formula of the moments
inertia, then they remain valid for a thin solid disk and
Symmetry (T.<
R.
z T.
Due to the smallness of the thickness of the cylinder
We believe that all points are
At the same distance R to the axis
And integration is not required.
Volume V \u003d 2πRTH. (Thin cylinder
Radius R with wall thickness T).
H.
Y.
X.
R.
I z R 2 DM R 2 2 RDRH
■
Z.
2
M ((R 2 (R T) 2) M (2 R 2 2 RT T 2) 2R.
IZ.
.
2
2
We highlight the discrete small mass volume of Mi:
Δk zi Hi ΔMi VI Hi Δmi z hi z hi2 Δmi.
Z.
HI
I z R 2 2 RTH MR 2.
The same can be obtained using
Formulas for a thick-walled cylinder, given
A little t:
Kinetic moment of solid body
ΔMi.
X.
K z Δk zi z hi2 Δmi z i z.
VI
Or moving to infinitely small:
Y.
DK Z HDMV HDM Z H z H 2 DM.
K z dk z z h 2 dm z i z.
The kinetic moment of the rotating body is equal to the work
angular velocity at the moment of inertia relative to the axis of rotation.
22Theorem Euler
Theorems: Application of the amount of quantity amendment theorem
Movement of the system to the movement of a solid medium (water).
(x): m s (v2 x v1x) RxB RXs;
(y): m s (v2 y v1 y) R yob R ypov;
(z): Mreadtop
RZ is located
.
sec (v2 z v1 bill
z) RZVoda,
1. Select as an object
In the curvilinear channel of the turbine:
2. Return communication and replace them with reactions (RPOs - asylum surface forces)
3. Add active forces (RAB - automatic volumetric forces):
about
v1.
F1.
A.
A.
B.
B.
Rob
The amount of water movement at times T0 and T1
In the projectionshkak
Amount:
Axis:
Imagine
Q q.
0
C.
D.
F2.
v2.
AB
BC.
Q1 QBC QCD.
,
Changing the amount of water movement in the time interval:
Q Q1 QCD QAB.
Changing number
Movement
water vectors of the second amounts of fluid movement on the axis are equal
Difference
Projects
DQ DQCD DQAB, where DQAB (F1V1DT) V1;
For infinitely small
interval
of time
DT: vectors
The amount of projections of the main
Volumetric and surface forces on the same axis.
DQCD (F2V2 DT) V2.
Taking the product of density, cross-sectional area and speed per second mass
We get:
DQ (M DT) V;
AB
DQ.
RB RPov.
dt.
4. Record the theorem about changing the number of system movement:
Rpov
C.
D.
Pon
sec
1
DQCD (M secand DT) v2.
DQ m sec (v2 v1) dt.
M sec f1v1 f2v2,
Substituting the differential of the number of system movement
In the Amendment theorem we get:
M s (v2 v1) RB RPov.
The geometric difference between the vectors of the second amounts of fluid movement is equal to
The sum of the main vectors of volumetric and surface forces.
(Mechanical Systems) - IV Option
1. The main equation of the dynamics of the material point, as is known, is expressed by the equation. Differential equations of motion of arbitrary dots of a non-free mechanical system according to two ways to divide forces can be recorded in two forms:
(1) where k \u003d 1, 2, 3, ..., n is the number of points of the material system.
where - the mass of the k-so point; - the radius of the vector of the k-so point is a given (active) force acting on the K-th one or the resultant all active forces acting on the K-th one. - the resultant forces of reactions of bonds acting on the K-th one; - the resultant internal forces acting on the K-th one; - Equality of external forces acting on the K-th one.
With the help of equations (1) and (2), you can strive to decide both the first and second tasks of the speakers. However, the solution of the second task of the dynamics for the system is very complicated not only from a mathematical point of view, but also because we are faced with fundamental difficulties. They consist in the fact that both for the system (1) and for the system (2) the number of equations is significantly less than the number of unknown.
So, if used (1), then known for the second (reverse) task of the speakers will be and, and unknowns will be. The vector equations will be " n.", And unknown -" 2n ".
If proceed from the system of equations (2), then known and part of the external forces. Why part? The fact is that the external strength includes external reactions of connections that are unknown. In addition, unknowns will also be.
Thus, as the system (1) and the system (2) is unlocked. It is necessary to add equations, given the equations of links and it is possible to impose some restrictions on the connections themselves. What to do?
If we proceed from (1), then you can go along the way to compile the equations of the first kind of Lagrange. But this path is not rational because the more simpler task (less degrees of freedom), the harder from the point of view of mathematics to solve it.
Then we pay attention to the system (2), where - are always unknown. The first step when solving the system is to eliminate these unknowns. It should be borne in mind that we are usually not interested in the internal forces when the system is moving, that is, when the system is moving, you do not need to know how each point of the system is moving, but it is enough to know how the system is in general.
Thus, if unknown forces are excluded from the system (2) in various ways, we obtain some relations, that is, some common characteristics appear for the system, the knowledge of which make it possible to judge how the system is moving in general. These characteristics are entered by the so-called common speakers. Four such theorems:
1. Theorem O. moving Center Mass Mechanical System;
2. Theorem Ob. changing the number of mechanical system movement;
3. Theorem Ob. change the kinetic moment of the mechanical system;
4. Theorem Ob. change the kinetic energy of the mechanical system.
Ministry of Education and Science of the Russian Federation
Federal State Budgetary Educational Institution of Higher Professional Education
"Kuban State Technological University"
Theoretical mechanics
Part 2 Dynamics
Approved by the editorial publishing
university Council as
tutorial
Krasnodar
UDC 531.1 / 3 (075)
Theoretical mechanics. Part 2. Dynamics: Tutorial / L.I.Dyko; Kuban. State Tekhnol.Un-t. Krasnodar, 2011. 123 p.
ISBN 5-230-06865-5
Presents in brief form theoretical material, examples of solving problems are given, most of which reflect the actual issues of technology, attention is paid to the choice of a rational method of solution.
Designed for bachelors by correspondence and remote forms of training, transport and engineering directions.
Table. 1 ill. 68 Bibliogr. 20 Names.
Scientific editor Cand.Tehn. Nauk, Assoc. V.F. Melnikov
Reviewers: Department of theoretical mechanics and theory of mechanisms and machines of the Kuban Agrarian University of Prof. FM Canarev; Associate Professor of the Department of Theoretical Mechanics of the Kuban State Technological University M.E. Mults
Printed by the decision of the editorial and publishing council of the Kuban State Technological University.
Reprint
ISBN 5-230-06865-5 Kubbda 1998.
Preface
This tutorial is intended for students of correspondence formation of construction, transport and engineering specialties, but it can be used in the study of the section "Dynamics" of the theoretical mechanics of students with students of other specialties, as well as students of the day form of training in independent work.
The manual is drawn up in accordance with the current program of the theoretical mechanics, covers all the issues of the main part of the course. Each section contains a short theoretical material, equipped with illustrations and methodical guidelines for its use when solving tasks. The manual disassembled the decision of 30 tasks reflecting the real issues of technology and the corresponding control tasks for an independent decision. For each task, a design scheme is presented, a clearly illustrating solution. The decisive of the decision complies with the requirements for the registration of control work of joys students.
The author expresses deep appreciation to teachers of the department of theoretical mechanics and the theory of mechanisms and machines of the Kuban Agrarian University for a lot of work on reviewing the textbook, as well as teachers of the Department of Theoretical Mechanics of the Kuban State Technological University for valuable comments and tips on the preparation of textbooks for publication.
All critical comments and wishes will be accepted by the author with gratitude and later.
Introduction
Dynamics is the most important section of theoretical mechanics. Most specific tasks that are in engineering practice belong to the dynamics. Using the conclusions of statics and kinematics, the dynamics establishes the general laws of motion of material bodies under the action of the applied forces.
The simplest material object is the material point. For a material point, you can take the material body of any shape, the sizes of which in the problem under consideration can be neglected. For a material point, you can take the body of the final sizes if the difference in the movement of its points for this task is not significant. This happens when the sizes of the body are small compared to distances that pass the body points. Every particle of solid can be considered a material point.
Forces attached to the point or material body, in dynamics are estimated at their dynamic effect, that is, according to how they change the characteristics of the motion of material objects.
The movement of material objects over time is performed in space relative to a certain reference system. In classical mechanics based on Newton's axioms, the space is considered three-dimensional, its properties do not depend on the material objects moving in it. The position of the point in such space is determined by three coordinates. Time is not related to the space and movement of material objects. It is considered the same for all reference systems.
The laws of speakers describe the movement of material objects in relation to the absolute axes of coordinates, conditionally adopted for fixed. The beginning of the absolute coordinate system is accepted in the center of the Sun, and the axes are sent to remote, conditionally non-moving stars. When solving many technical tasks, the coordinate axes associated with the Earth can be considered conditionally movable.
The parameters of the mechanical movement of material objects in the dynamics are established by mathematical conclusions from the basic laws of classical mechanics.
First Law (Law of Inertia):
The material point retains the state of rest or uniform and rectilinear movement until the action of any forces will displays it from this state.
Uniform and rectilinear movement of the point are called inertia movement. Pochka is a special case of inertia, when the speed of the point is zero.
Any material point has inertia, i.e., seeks to preserve the state of rest or uniform rectilinear movement. The reference system, with respect to which the law of inertia is performed, is called inertial, and the movement observed with respect to this system is called absolute. Any reference system that performs relative to the inertial system translational straight and uniform movement will also be an inertial system.
Second law (basic law of dynamics):
The acceleration of the material point relative to the inertial reference system is proportional to the force attached to the point and coincides with the force towards: 
.
Of the basic law, the dynamics follows that in force 
acceleration 
. The mass of the point characterizes the degree of resistance to the point of change in its speed, i.e. is a measure of the inertness of the material point.
Third Law (law of action and counteraction):
Forces with which two bodies act on each other are equal to the module and are directed along one straight to the opposite sides.
Forces, referred to as the action and opposition, are applied to different bodies and therefore the balanced system does not form.
Fourth Law (the law of independence of the strength):
With a simultaneous action of several forces, the acceleration of the material point is equal to the geometric amount of accelerations that would have a point under the action of each force separately:
where 
,
,…,
.
Consider the movement of some material volume system with respect to the fixed coordinate system when the system is not free, it can be viewed as free if you drop the connection superimposed on the system and replace them with appropriate reactions.
We divide all the forces attached to the system, to external and internal; In those and others can enter the reactions of discarded
connections. Through and denote the main vector and the main moment of external forces relative to the point A.
1. The theorem on the change in the amount of movement. If - the number of system movement, then (see)
i.e., the theorem is true: time derivative on the amount of system movement is equal to the main vector of all external forces.
Replacing the vector through its expression where - the mass system - the speed of the center of mass, equation (4.1) can be given another form:
This equality means that the center of the mass system moves, as the material point of the mass of which is equal to the mass of the system and to which the force is applied, geometrically equal to the main vector of all external system forces. The last statement is called the theorem on the movement of the center of the mass (center of the inertia) of the system.
If then from (4.1) it follows that the vector of the amount of movement is constant in magnitude and direction. Designing it on the axis of the coordinates, we obtain three scalar first integrals, differential equations of the DVNsckeeping of the system:
These integrals are called the integrals of the amount of movement. With the speed of the center of mass is constant, i.e. it moves evenly and straightly.
If the projection of the main vector of external forces on any one axis, for example, on the axis is zero, then we have one first integral or even equal to zero "two projections of the main vector, then there are two integral of the amount of movement.
2. The theorem on the change in the kinetic moment. Let a be some arbitrary point of space (moving or fixed), which does not necessarily coincide with any particular material point of the system at all time movement. Its speed in the fixed sprochene coordinates denote through the theorem on the change in the kinetic moment of the material system relative to the point A has the form
If the point A is stationary, then equality (4.3) takes a simpler form:
This equality expresses the theorem on the cinetic moment of the system relative to the fixed point: the time derivative from the kinetic moment of the system calculated relative to a certain fixed point is equal to the main point of all external forces relative to this point.
If then according to (4.4), the kinetic moment vector is constant largest and direction. Designing it on the axis of the coordinates, we obtain the scalar first integral of the differential equations of the District System:
These integrals will appear the name of the integrals of the kinetic moment or integrals of space.
If the point A coincides with the center of the mass system, then the first term in the right-hand side of equality (4.3) refers to zero and the theorem on the change in the kinetic moment has the same form of record (4.4), which in the case of a fixed point A. We note (see § 4 § 3) that in the case under consideration the absolute kinetic moment of the system in the left part of equality (4.4) can be replaced by the kinetic moment of the system in its movement relative to the center of mass.
Let - be some unchanging axis of the axis of the unchanged direction passing through the center of the mass of the system, and the kinetic moment of the system relative to this axis. From (4.4) it follows that
where is the moment of external forces regarding the axis. If at all time movement then we have the first integral
In the works of S. A. Chaplygin, several generalizations of the theorem on the change in the kinetic moment were obtained, which were then applied when solving a number of tasks about rolling balls. Further generalizations of the theorem on the change in the CPNTCH momentum and their applications in the tasks of the DNNNAMIC solid body are contained in the works. The main results of these works are associated with the theorem on the change in the kinetic moment relative to the movable, constantly passing through a somewhat moving point. A. Let a single vector directed along this axis. Multiplying is scalaring on both parts of equality (4.3) and adding parts to its discourse
When performing the kinematic condition
from (4.7), equation (4.5) follows. And if at all the time of movement and condition (4.8) is satisfied, then there is the first integral (4.6).
If the links of the system are ideal and allowed among the virtual displacements of the rotation of the system as a solid body around the axis and, then the main moment of the reactions relative to the axis is zero, and then the value in the right-hand side of equation (4.5) is the main point of all external active forces relative to the axis and . Equality zero of this moment and the feasibility of the relation (4.8) will be in the case under consideration sufficient conditions for the existence of the integral (4.6).
If the direction of the axis and the invariably condition (4.8) is recorded as
This equality means that the projections of the velocity of the center of mass and the point of point and the axis and the plane perpendicular to this are parallel. In the work of S. A. Chaplygin instead of (4.9), it is necessary to perform a less general condition where X is an arbitrary constant value.
Note that condition (4.8) does not depend on the choice of point on. Indeed, let P be an arbitrary point on the axis. Then
and, therefore,
In conclusion, we note the geometric interpretation of the cut of equations (4.1) and (4.4): the vectors of the absolute speeds of the ends of the vectors and are equal to the co-external main vector and the main point of all external forces relative to the point A.
