The amount of movement of the system and how it is calculated. The angular momentum of a material point relative to the center and axis. See what "Movement amount" is in other dictionaries
and mechanical system
The amount of movement of a material point is a vector measure of mechanical movement, equal to the product of the point's mass by its speed,. The unit of measurement of the amount of motion in the SI system - 
... The number of movements of a mechanical system is equal to the sum of the numbers of movements of all material points that form the system:
.
(5.2)
We transform the resulting formula
.
According to formula (4.2) 
, therefore
.
Thus, the momentum of a mechanical system is equal to the product of its mass by the speed of the center of mass:
.
(5.3)
Since the momentum of a system is determined by the motion of only one of its points (center of mass), it cannot be a complete characteristic of the motion of the system. Indeed, for any motion of the system, when its center of mass remains stationary, the momentum of the system is zero. For example, this occurs when a rigid body rotates around a fixed axis passing through its center of mass.
Let us introduce a frame of reference Cxyz originating in the center of mass of the mechanical system WITH and moving translationally relative to the inertial system 
(fig. 5.1). Then the movement of each point 
can be viewed as complex: a portable movement along with the axes Cxyz and movement about these axes. Due to the translational motion of the axes Cxyz the portable speed of each point is equal to the speed of the center of mass of the system, and the momentum of the system, determined by formula (5.3), characterizes only its translational translational motion.
5.3. Impulse of force
To characterize the action of a force over a certain period of time, a quantity called impulse of power ... An elementary impulse of force is a vector measure of the action of a force, equal to the product of a force by an elementary period of time of its action:
.
(5.4)
The SI unit of force impulse is 
, i.e. the dimensions of the impulse of force and momentum are the same.
Impulse of force in a finite period of time 
is equal to definite integral from an elementary impulse:
.
(5.5)
The impulse of constant force is equal to the product of force by the time of its action:
.
(5.6)
In the general case, the impulse of force can be determined from its projections on the coordinate axes:
.
(5.7)
5.4. Momentum Change Theorem
material point
In the basic equation of dynamics (1.2), the mass of a material point is a constant value, its acceleration 
, which makes it possible to write this equation in the form:
.
(5.8)
The resulting ratio allows us to formulate the theorem on the change in the amount of motion of a material point in differential form: The time derivative of the momentum of a material point is equal to the geometric sum (main vector) of the forces acting on the point.
Now we get the integral form of this theorem. It follows from relation (5.8) that
.
We integrate both sides of the equality within the limits corresponding to the moments of time 
and 
,
.
(5.9)
The integrals on the right side represent the impulses of forces acting on the point, therefore, after integrating the left side, we obtain
.
(5.10)
Thus, it is proved theorem on the change in the amount of motion of a material point in integral form: The change in the amount of motion of a material point over a certain period of time is equal to the geometric sum of the impulses of forces acting on the point for the same period of time.
The vector equation (5.10) corresponds to a system of three equations in projections onto the coordinate axes:
;
;
(5.11)
.
Example 1.
The body moves translationally along an inclined plane making an angle α with the horizon. At the initial moment of time, it had a speed 
directed upward along an inclined plane (Fig. 5.2).
After what time does the body's speed become equal to zero if the coefficient of friction is f ?
Let us take a translationally moving body as a material point and consider the forces acting on it. It's gravity 
, the normal reaction of the plane 
and friction force 
... Let's direct the axis x along the inclined plane upwards and write down the 1st equation of the system (5.11)
where the projections of the quantities of motion, and the projections of the impulses of constant forces 
,
and 
are equal to the products of the projections of forces for the time of movement:
Since the acceleration of the body is directed along the inclined plane, the sum of the projections onto the axis y of all forces acting on the body is equal to zero: 
, whence it follows that 
... Find the friction force
and from equation (5.12) we obtain
from where we determine the time of body movement
.
The amount of movement of the system is called the geometric sum of the quantities of movement of all material points of the system
To clarify the physical meaning of (70), we calculate the derivative of (64)
.
(71)
Solving (70) and (71) together, we obtain
.
(72)
Thus, the vector of the momentum of a mechanical system is determined by the product of the mass of the system by the speed of its center of mass.
Let us calculate the derivative of (72)
.
(73)
Solving (73) and (67) together, we obtain
.
(74)
Equation (74) expresses the following theorem.
Theorem: The time derivative of the system's angular momentum vector is equal to the geometric sum of all external forces of the system.
When solving problems, equation (74) must be designed on the coordinate axes:
.
(75)
The analysis of (74) and (75) implies the following. law of conservation of momentum of a system: If the sum of all the forces of the system is equal to zero, then the vector of its momentum retains its magnitude and direction.
If 
, then 
,Q
=
const
.
(76)
In a particular case, this law can be fulfilled along one of the coordinate axes.
If 
, then, Q z =
const.
(77)
It is advisable to use the theorem on the change in the momentum when liquid and gaseous bodies are included in the system.
Theorem on the change in the angular momentum of a mechanical system
The amount of movement characterizes only the translational component of the movement. To characterize the rotational motion of the body, the concept of the main moment of the quantities of motion of the system relative to the this center (kinetic moment).
Kinetic moment of the system relative to a given center is called the geometric sum of the moments of the quantities of motion of all its points relative to the same center
.
(78)
By projecting (22) on the coordinate axis, one can obtain the expression for the angular momentum relative to the coordinate axes
.
(79)
Kinetic moment of the body relative to the axes is equal to the product of the moment of inertia of the body about this axis by the angular velocity of the body
.
(80)
From (80) it follows that the angular momentum characterizes only the rotational component of the motion.
The characteristic of the rotational action of a force is its moment relative to the axis of rotation.
The theorem on the change in angular momentum establishes the relationship between the characteristic of rotational motion and the force causing this motion.
Theorem: The time derivative of the angular momentum vector of the system with respect to some center is equal to the geometric sum of the moments of all external forces of the system with respect tothe same center
.
(81)
When solving engineering problems (81), it is necessary to design on the coordinate axes
Their analysis (81) and (82) implies momentum conservation law: If the sum of the moments of all external forces relative to the center (or axis) is zero, then the angular momentum of the system relative to this center (or axis) retains its magnitude and direction.
,
or 
The kinetic moment cannot be changed by the action of the internal forces of the system, but due to these forces it is possible to change the moment of inertia, and hence the angular velocity.
- 1. Algebraic angular momentum relative to the center. Algebraic O- scalar value, taken with the (+) or (-) sign and equal to the product of the modulus of the momentum m at a distance h(perpendicular) from this center to the line along which the vector is directed m:
- 2. Vector angular momentum relative to the center.
Vector angular momentum of a material point relative to some center O -- vector applied at this center and directed perpendicular to the plane of vectors m and in the direction from which the movement of the point is seen counterclockwise. This definition satisfies the vector equality
Momentum moment material point about some axis z is a scalar value taken with a (+) or (-) sign and equal to the product of the modulus vector projection amount of motion per plane perpendicular to this axis, perpendicular h, lowered from the point of intersection of the axis with the plane to the line along which the specified projection is directed:
The kinetic moment of the mechanical system about the center and axis
1. Kinetic moment relative to the center.
Kinetic moment or the main point of the quantities of motion of a mechanical system relative to some center is called the geometric sum of the moments of the quantities of motion of all material points of the system relative to the same center.
2. Kinetic moment about the axis.
The kinetic moment or the main moment of the quantities of motion of a mechanical system relative to some axis is called the algebraic sum of the moments of the quantities of motion of all material points of the system relative to the same axis.
3. The kinetic moment of a rigid body rotating around a fixed axis z with angular velocity.
Theorem on the change in the angular momentum of a material point relative to the center and axis
1. Theorem of moments about the center.
Derivative in time from the moment of momentum of a material point relative to some fixed center is equal to the moment of force acting on the point, relative to the same center
2. Theorem of moments about the axis.
Derivative in time from the moment of momentum of a material point relative to some axis is equal to the moment of force acting on the point, relative to the same axis
Theorem on the change in the angular momentum of a mechanical system relative to the center and axis
Center moment theorem.
Derivative in time from the angular momentum of the mechanical system relative to some fixed center is equal to the geometric sum of the moments of all external forces acting on the system, relative to the same center;
Consequence. If main point external forces relative to some center is equal to zero, then the angular momentum of the system relative to this center does not change (the law of conservation of angular momentum).
2. Theorem of moments about the axis.
Derivative in time from the angular momentum of the mechanical system relative to some fixed axis is equal to the sum of the moments of all external forces acting on the system, relative to this axis
Consequence. If the main moment of external forces relative to some axis is equal to zero, then the angular momentum of the system relative to this axis does not change.
For example, = 0, then L z = const.
Work and power of forces
Work of force is a scalar measure of the action of a force.
1. Elementary work of power.
Elementary the work of force is an infinitesimal scalar quantity equal to the scalar product of the force vector by the vector of infinite small displacement of the point of application of the force: ; - radius vector increment points of force application, the hodograph of which is the trajectory of this point. Elementary movement points along the trajectory coincides with due to their smallness. That's why
if then dA> 0; if, then dA = 0; if , then dA< 0.
2. Analytical expression of elementary work.
Let's represent vectors and d through their projections on the axis of Cartesian coordinates:
, . We get (4.40)
3. The work of force on a final displacement is equal to the integral sum of elementary work on this displacement
If the force is constant, and the point of its application moves in a straight line,
4. The work of gravity. We use the formula: Fx = Fy = 0; Fz = -G = -mg;
where h- moving the point of application of force vertically downward (height).
When moving the point of application of gravity up A 12 = -mgh(point M 1 -- at the bottom, M 2 - top).
So, . The work of gravity does not depend on the shape of the trajectory. When moving along a closed trajectory ( M 2 matches M 1 ) work is zero.
5. The work of the spring force.
The spring only stretches along the axis NS:
F y = F z = O, F x = = -cx;
where is the amount of spring deformation.
When moving the point of application of force from the lower position to the upper position, the direction of the force and the direction of movement coincide, then
Therefore, the work of the elastic force
The work of forces on the final displacement; If = const, then
where is the final angle of rotation; , where NS -- the number of revolutions of the body around the axis.
Kinetic energy of a material point and a mechanical system. Koenig's theorem
Kinetic energy is a scalar measure of mechanical motion.
The kinetic energy of a material point is a scalar positive value equal to half the product of the point's mass by the square of its velocity,
The kinetic energy of a mechanical system is the arithmetic sum of the kinetic energies of all material points of this system:
The kinetic energy of a system consisting of NS of interconnected bodies is equal to the arithmetic sum of the kinetic energies of all bodies of this system:
Koenig's theorem
Kinetic energy of a mechanical system in the general case, its motion is equal to the sum of the kinetic energy of motion of the system together with the center of mass and the kinetic energy of the system when it moves relative to the center of mass:
where Vkc - speed k- th points of the system relative to the center of mass.
Kinetic energy of a rigid body with different motion
Translational motion.
Rotation of a body around a fixed axis ... ,where -- moment of inertia of the body about the axis of rotation.
3. Plane-parallel movement. , where is the moment of inertia of a plane figure relative to the axis passing through the center of mass.
When moving flat body kinetic energy consists of the kinetic energy of the translational motion of the body with the speed of the center of mass and kinetic energy of rotational motion about an axis passing through the center of mass;
The theorem on the change in the kinetic energy of a material point
Theorem in differential form.
Differential of the kinetic energy of a material point is equal to the elementary work of the force acting on the point,
The theorem in integral (final) form.
The change the kinetic energy of a material point at a certain displacement is equal to the work of the force acting on the point at the same displacement.
The theorem on the change in the kinetic energy of a mechanical system
Theorem in differential form.
Differential from the kinetic energy of the mechanical system is equal to the sum elementary work of external and internal forces acting on the system.
The theorem in integral (final) form.
The change the kinetic energy of a mechanical system at a certain displacement is equal to the sum of the work of external and internal forces applied to the system at the same displacement. ; For a system of rigid bodies = 0 (by the property of internal forces). Then
The law of conservation of mechanical energy of a material point and a mechanical system
If the material point or mechanical system only conservative forces act, then in any position of the point or system the sum of kinetic and potential energies remains constant.
For a material point
For mechanical system T + P = const
where T + P - complete mechanical energy systems.
Rigid body dynamics
Differential equations of motion for a rigid body
These equations can be obtained from general theorems of the dynamics of a mechanical system.
1. The equations of translational motion of a body - from the theorem on the motion of the center of mass of a mechanical system In projections on the axis of Cartesian coordinates
2. The equation of rotation of a rigid body about a fixed axis - from the theorem on the change in the angular momentum of a mechanical system about an axis, for example, about an axis
Since the kinetic moment L z of a rigid body relative to the axis, then if
Since either, the equation can be written in the form or, the form of the equation depends on what should be determined in a particular problem.
Differential equations of plane-parallel the motion of a rigid body is a set of equations progressive the movement of a flat figure together with the center of mass and rotational motion about an axis passing through the center of mass:
Physical pendulum
Physical pendulum is called a rigid body rotating around a horizontal axis that does not pass through the center of mass of the body, and moving under the action of gravity.
Differential Equation of Rotation
In the case of small fluctuations.
Then where
Solution to this homogeneous equation.
Let at t = 0 Then
-- the equation of harmonic vibrations.
Pendulum swing period
Reduced length a physical pendulum is the length of such a mathematical pendulum, the oscillation period of which is equal to the oscillation period of the physical pendulum.
The amount of motion of the system, as a vector quantity, is determined by formulas (4.12) and (4.13).
Theorem. The time derivative of the system's momentum is equal to the geometric sum of all external forces acting on it.
In the projections of the Cartesian axes, we obtain scalar equations.
You can write a vector
(4.28)
and scalar equations
Which express the theorem 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 impulses over the same period of time. When solving problems, equations (4.27) are often used
The law of conservation of momentum
The theorem on the change in the 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 stationary center is equal to the vector moment acting on the point of force relative to the same center.
or 
(4.30)
Comparing (4.23) and (4.30), we see that the moments of the vectors and are related by the same dependence as the vectors and are related (Fig. 4.1). If we project the equality onto the axis passing through the center O, then we get
(4.31)
This equality expresses the theorem of the angular momentum of a point about an axis.
|
(4.32)
If we project expression (4.32) on the axis passing through the center O, then 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 differential equation a rotating solid (wheels, axles, shafts, rotors, etc.) in three forms.
(4.34)
(4.35)
(4.36)
Thus, it is advisable to use the theorem on the change in the angular momentum to study the motion of a rigid body, which is very widespread in technology, and its rotation around a fixed axis.
The law of conservation of the angular momentum of the system
1. Let in expression (4.32).
Then it follows from equation (4.32) that, i.e. if the sum of the moments of all external forces applied to the system relative to a given center is equal to zero, then the kinetic moment of the system relative to this center will be numerically and will be constant in direction.
2. If, then. Thus, if the sum of the moments of external forces acting on the system relative to a certain axis is equal to zero, then the angular momentum of the system relative to this axis will be constant.
These results express the law of conservation of angular momentum.
In the case of a rotating rigid body, equality (4.34) implies that, if, then. From here we come to the following conclusions:
If the system is unchangeable (absolutely rigid body), then, consequently, the rigid body also rotates around a fixed axis with a constant angular velocity.
If the system is changeable, then. With increasing (then individual elements of the system move away from the axis of rotation), the angular velocity decreases, because , and with a decrease it increases, thus, in the case of a variable system, using internal forces, it is possible to change the angular velocity.
Second task D2 test work is devoted to the theorem on the change in the angular momentum of the system with respect to the axis.
Problem D2
A uniform horizontal platform (round 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.D2.0 - D2.9, table D2); dimensions for all rectangular platforms are shown in fig. D2.0a (top view).
At the moment of time, a load D with a mass of kg begins to move along the platform chute (under the action of internal forces) according to the law, where s is expressed in meters, t - in seconds. Simultaneously, a pair of forces begins to act on the platforms with the moment M (given in newtonometers; for M< 0 его направление противоположно показанному на рисунках).
Determine, neglecting the mass of the shaft, the dependence, i.e. the angular velocity of the platform as a function of time.
In all the figures, the load D is shown in the position at 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 - on the application of the theorem on the change in the angular momentum of the system. When applying the theorem to a system consisting of a platform and a load, the angular momentum of the system relative to the z axis is determined as the sum of the moments of the platform and the load. It should be borne in mind that the absolute speed of the cargo is the sum of the relative and portable speeds, i.e. ... Therefore, the amount of movement of this cargo 
... Then you can use Varignon's theorem (statics), according to which; these moments are calculated in the same way as the moments of forces. The course of the solution is explained in more detail in example D2.
When solving the problem, it is useful to depict in the auxiliary drawing a view of the platform from above (from the end z), as is done in Fig. D2.0, a - D2.9, a.
The moment of inertia of a plate with mass m relative to the Cz axis, perpendicular to the plate and passing through its center of mass, is: for a rectangular plate with sides and
;
For round insert of radius R
| Condition number | b | s = F (t) | M |
| R R / 2 R R / 2 R R / 2 R R / 2 R R / 2 | -0.4 0.6 0.8 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), which has a mass, is rigidly attached to the vertical shaft and rotates with it around the axis z with angular velocity (Fig.D2a ). At the moment of time, the torque M begins to act on the shaft, directed oppositely ; simultaneously cargo D mass in the gutter AB at the point WITH, begins to move along the chute (under the action 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 with respect to the axis z:
(1)
Let us represent the external forces acting on the system: the gravity forces of the reaction and the torque M. Since the forces and are parallel to the z-axis, and the reactions also intersect this axis, their moments about the z-axis are equal to zero. Then, assuming the direction for the moment to be positive (i.e. counterclockwise), we obtain 
and equation (1) will take this form.
The amount of movement of a material point is called the vector quantity mV, equal to the product of the mass of a point by the vector of its velocity. Vector mV attached to a moving point.
The amount of movement of the system is called the vector quantity Q equal to the geometric sum (principal vector) of the quantities of motion of all points of the system:
Vector Q is a free vector. In SI units, the momentum modulus is measured in kg m / s or N s.
As a rule, the velocities of all points of the system are different (see, for example, the distribution of the velocities of the points of a rolling wheel shown in Fig. 6.21), and therefore the direct summation of the vectors on the right-hand side of equality (17.2) is difficult. Let us find a formula with which the value Q it is much easier to calculate. It follows from equality (16.4) that
Taking the time derivative from both sides, we get
Hence, taking into account equality (17.2), we find that
that is, the momentum of the system is equal to the product of the mass of the entire system by the speed of its center of mass.
Note that the vector Q, like the main vector of forces in statics, it is some generalized vector characteristic of the motion of the entire mechanical system. In the general case of movement of the system, its momentum Q can be considered as a characteristic of the translational part of the motion of the system together with its center of mass. If during the movement of the system (body) the center of mass is motionless, then the momentum of the system will be equal to zero. This is, for example, the amount of motion of a body rotating around a fixed axis passing through its center of mass.
Example. Determine the amount of movement of the mechanical system (Fig.17.1, a), consisting of cargo A mass t A - 2 kg, homogeneous block V weighing 1 kg and wheels D mass m D - 4 kg. Cargo A moves with speed V A - 2 m / s, wheel D rolls without slipping, the thread is inextensible and weightless. Solution. The amount of movement of the system of bodies
Body A moves progressively and Q A = m A V A(numerically Q A= 4 kg m / s, vector direction Q A coincides with the direction V A). Block V commits rotary motion around a fixed axis passing through its center of mass; hence, Q B - 0. Wheel D performs a plane-parallel
traffic; its instantaneous center of velocities is at the point TO, therefore, the speed of its center of mass (points E) is equal to V E = V A / 2 = 1 m / s. Wheel movement amount Q D - m D V E - 4 kg m / s; vector Q D directed horizontally to the left.
By drawing vectors Q A and Q D in fig. 17.1, b, we find the amount of motion Q systems according to formula (a). Taking into account the directions and numerical values of the quantities, we get Q ~ ^ Q A + Q E= 4l / 2 ~ kg m / s, vector direction Q shown in fig. 17.1, b.
Considering that a -dV / dt, equation (13.4) of the basic law of dynamics can be represented as
Equation (17.4) expresses the theorem on the change in the momentum of a point in differential form: at each moment of time, the time derivative of the momentum of the point is equal to the force acting on the point. (In essence, this is another formulation of the basic law of dynamics, close to the one given by Newton.) If several forces act on a point, then on the right side of equality (17.4) there will be a resultant of forces applied to a material point.
If both sides of the equality are multiplied by dt, we get
The vector quantity on the right side of this equality characterizes the action exerted on the body by the force for an elementary period of time dt this value is denoted dS and called an elementary impulse of force, i.e.
Pulse S strength F over a finite time interval /, - / 0 is defined as the limit of the integral sum of the corresponding elementary impulses, i.e.
In a particular case, if the force F constant modulo and direction, then S = F (t| - / 0) and S- F (t l -/ 0). In the general case, the modulus of a force impulse can be calculated from its projections onto the coordinate axes:
Now, integrating both sides of equality (17.5) for T= const, we get
Equation (17.9) expresses the theorem on the change in the momentum of a point in the final (integral) form: the change in the momentum of a point for a certain period of time is equal to the impulse of the force acting on the point (or the impulse of the resultant of all forces applied to it) for the same period of time.
When solving problems, the equations of this theorem are used in projections on the coordinate axes
Now consider a mechanical system consisting of NS material points. Then, for each point, the theorem on the change in the momentum in the form (17.4) can be applied, taking into account the external and internal forces applied to the points:
Summing these equalities and taking into account that the sum of the derivatives is equal to the derivative of the sum, we obtain
Since by the property of internal forces HF k= 0 and by definition of the amount of motion ^ fn k V / c = Q, then we finally find
Equation (17.11) expresses the theorem on the change in the momentum of the system in differential form: at each moment of time, the time derivative of the momentum of the system is equal to the geometric sum of all external forces acting on the system.
Projecting equality (17.11) onto the coordinate axes, we obtain
Multiplying both sides of (17.11) by dt and integrating, we get
where 0, Q 0 - the momentum of the system at the moments of time, respectively, and / 0.
Equation (17.13) expresses the theorem on the change in the momentum of the system in integral form: the change in the momentum of the system for any time is equal to the sum of the impulses of all external forces acting on the system during the same time.
In projections onto the coordinate axes, we obtain
From the theorem on the change in the momentum of the system, the following important consequences can be obtained, which express the law of conservation of the momentum of the system.
- 1. If the geometric ^ umma of all external forces acting on the system is zero (LF k= 0), then it follows from equation (17.11) that in this case Q= const, i.e. the vector of the momentum of the system will be constant in magnitude and direction.
- 2. If the external forces acting on the system are such that the sum of their projections onto any axis is zero (for example, I e kx = 0), then it follows from equations (17.12) that in this case Q x = const, i.e. the projection of the momentum of the system on this axis remains unchanged.
Note that the internal forces of the system do not participate in the equation of the theorem about the change in the momentum of the system. These forces, although they affect the amount of movement of individual points of the system, cannot change the amount of movement of the system as a whole. Considering this circumstance, when solving problems, it is advisable to choose the system under consideration so that the unknown forces (all or part of them) are made internal.
The law of conservation of momentum is convenient to apply in cases when the speed of another part of the system must be determined from the change in the speed of one part of the system.
Task 17.1. TO trolley weight t x- 12 kg moving on a smooth horizontal plane at a point A a weightless rod is attached using a cylindrical hinge AD length / = 0.6 m with load D mass t 2 - 6 kg at the end (fig.17.2). At the moment of time / 0 = 0, when the trolley speed and () - 0.5 m / s, rod AD starts to rotate around the axis A, perpendicular to the plane of the drawing, according to the law φ = (n / 6) (3 ^ 2 - 1) rad (/ -in seconds). Define: u = f.
§ 17.3. The theorem on the motion of the center of mass
The theorem on the change in the momentum of a mechanical system can be expressed in another form, which is called the theorem on the motion of the center of mass.
Substituting into equation (17.11) the equality Q = MV C, get
If the mass M system is constant, then we get
where and with - acceleration of the center of mass of the system.
Equation (17.15) also expresses the theorem on the motion of the center of mass of the system: the product of the mass of the system and the acceleration of its center of mass is equal to the geometric sum of all external forces acting on the system.
Projecting equality (17.15) onto the coordinate axes, we obtain
where x c, y c, z c - coordinates of the center of mass of the system.
These equations are differential equations of motion of the center of mass in projections on the axis of the Cartesian coordinate system.
Let us discuss the results obtained. Let us first recall that the center of mass of the system is a geometric point located sometimes outside the geometric boundaries of the body. The forces acting on the mechanical system (external and internal) are applied to all material points of the system. Equations (17.15) make it possible to determine the motion of the center of mass of the system without determining the motion of its individual points. Comparing equations (17.15) of the theorem on the motion of the center of mass and equations (13.5) of Newton's second law for a material point, we come to the conclusion: the center of mass of a mechanical system moves like a material point, the mass of which is equal to the mass of the entire system, and as if all external forces acting on the system are applied to this point. Thus, the solutions that we obtain, considering a given body as a material point, determine the law of motion of the center of mass of this body.
In particular, if the body is moving translationally, then the kinematic characteristics of all points of the body and its center of mass are the same. That's why a translationally moving body can always be considered as a material point with mass, equal mass of the whole body.
As can be seen from (17.15), the internal forces acting on the points of the system do not affect the motion of the center of mass of the system. Internal forces can affect the movement of the center of mass in cases where external forces change under their influence. Examples of this will be given below.
From the theorem on the motion of the center of mass, the following important consequences can be obtained, which express the conservation law of the motion of the center of mass of the system.
1. If the geometric sum of all external forces acting on the system is zero (LF k= 0), then from equation (17.15) it follows that
that at the same time and c = 0 or V c = const, i.e., the center of mass of this system
moves with a speed constant in absolute value and direction (otherwise, uniformly and rectilinearly). In a particular case, if at first the center of mass was at rest ( V c= 0), then he will remain at rest; where
track he knows that his position in space will not change, i.e. r c = const.
2. If the external forces acting on the system are such that the sum of their projections onto some axis (for example, the axis NS) is zero (? F e kx= 0), then it follows from equation (17.16) that in this case x with= 0 or V Cx = x c = const, that is, the projection of the velocity of the center of mass of the system onto this axis is a constant value. In a particular case, if at the initial moment Vex= 0, then at any subsequent moment of time this value will be preserved, and from this it follows that the coordinate x with the center of mass of the system will not change, i.e. x c - const.
Let's consider examples illustrating the law of motion of the center of mass.
Examples. 1. As noted, the movement of the center of mass depends only on external forces; it is impossible to change the position of the center of mass by internal forces. But the internal forces of the system can cause external influences. So, the movement of a person on a horizontal surface occurs under the action of frictional forces between the soles of his shoes and the surface of the road. The force of his muscles (internal forces) pushes a person with his feet from the surface of the road, which is why a friction force arises at the points of contact with the road (external for a person), directed in the direction of his movement.
- 2. The car moves in the same way. Internal pressure forces in its engine make the wheels rotate, but since the latter have adhesion to the road, the resulting frictional forces "push" the car forward (as a result, the wheels do not rotate, but move in a plane-parallel manner). If the road is absolutely smooth, then the center of mass of the car will be stationary (at zero initial speed) and the wheels, in the absence of friction, will slip, that is, make a rotational motion.
- 3. Movement with the help of a propeller, propeller, oars occurs due to the rejection of a certain mass of air (or water). If we consider the discarded mass and the moving body as one system, then the forces of interaction between them, as internal, cannot change the total amount of motion of this system. However, each of the parts of this system will move, for example, the boat forward, and the water, which is thrown by the oars, - back.
- 4. In an airless space, when the rocket moves, the "discarded mass" should be "taken with you": the jet engine imparts motion to the rocket by throwing back the combustion products of the fuel with which the rocket is fueled.
- 5. When descending by parachute, it is possible to control the movement of the center of mass of the man-parachute system. If, with muscular efforts, a person pulls the parachute lines so that the shape of his canopy or the angle of attack of the air flow changes, this will cause a change in the external influence of the air flow, and thereby influence the movement of the entire system.
Task 17.2. V To task 17.1 (see fig. 17.2) determine: 1) trolley motion law NS (= /) (/) if it is known that at the initial moment of time t 0 = The system was at rest and the coordinate x 10 = 0; 2) ^ the law of change with time of the total value of the normal reaction N (N = N "+ N") horizontal plane, i.e. N = f 2 (t).
Solution. Here, as in Problem 17.1, consider a system consisting of a cart and a load D, in an arbitrary position under the action of external forces applied to it (see Fig. 17.2). Coordinate Axes Ooh draw so that the x axis is horizontal, and the axis at went through the point A 0, i.e. the location of the point A at the moment t-t 0 - 0.
1. Determination of the law of motion of the trolley. To determine x, = /, (0, we use the theorem on the motion of the center of mass of the system. Let us compose the differential equation of its motion in projection onto the x-axis:
Since all external forces are vertical, then T, F e kx = 0, and therefore
Integrating this equation, we find that Mx c = B, that is, the projection of the velocity of the center of mass of the system onto the x-axis is a constant value. Since at the initial moment of time
By integrating the equation Mx c= 0, we get
i.e. coordinate x with the center of mass of the system is constant.
Let's write the expression Mx c for an arbitrary position of the system (see Fig. 17.2), taking into account that x A - x { , x D - x 2 and x 2 - x ( - I sin f. In accordance with formula (16.5), which determines the coordinate of the center of mass of the system, in this case Mx c - t (x ( + t 2 x 2 ".
for an arbitrary moment in time
for the moment of time / () = 0, NS (= 0 and
In accordance with equality (b), the coordinate x with the center of mass of the entire system remains unchanged, i.e., xD ^,) = x c (t). Therefore, equating expressions (c) and (d), we obtain the dependence of the x coordinate on time.
Answer: NS - 0.2 m, where t - in seconds.
2. Definition of the reaction N. For determining N = f 2 (t) we compose the differential equation of motion of the center of mass of the system in projection onto the vertical axis at(see fig.17.2):
Hence, denoting N = N + N ", get
According to the formula determining the ordinate with the center of mass of the system, Mu s = t (y x + t 2 y 2, where y = at C1,at 2= y D = Havea ~ 1 cos Ф "we get
Differentiating this equality twice in time (taking into account that at C1 and at A values are constant and, therefore, their derivatives are equal to zero), we find
Substituting this expression into equation (e), we determine the desired dependence N from t.
Answer: N- 176,4 + 1,13,
where φ = (π / 6) (3 / -1), t - in seconds, N- in newtons.
Task 17.3. Electric motor mass t x bolted to the horizontal surface of the foundation (fig. 17.3). On the motor shaft at right angles to the axis of rotation, a weightless rod of length / is fixed at one end, a point weight is mounted on the other end of the rod A mass t 2. The shaft rotates uniformly with an angular velocity c. Find the horizontal motor pressure on the bolts. Solution. Consider a mechanical system consisting of a motor and a point weight A, in an arbitrary position. Let us depict external forces acting on the system: gravity P x, P 2, foundation reaction in the form of vertical force N and horizontal force R. Let's draw the x-axis horizontally.
To determine the horizontal pressure of the motor on the bolts (and it will be numerically equal to the reaction R and directed opposite to the vector R ), we compose the equation of the theorem about the change in the momentum of the system in projection onto the horizontal x-axis:
For the system under consideration in its arbitrary position, given that the momentum of the motor body is zero, we get Q x = - t 2 U A com. Taking into account that V A = a s /, f = w / (the rotation of the motor is uniform), we get Q x - - m 2 co / cos co /. Differentiating Q x in time and substituting into equality (a), we find R- m 2 co 2 / sin co /.
Note that it is precisely such forces that are compelling (see § 14.3); when they act, forced vibrations of structures arise.
Exercises for independent work
- 1. What is called the amount of motion of a point and a mechanical system?
- 2. How does the amount of motion of a point uniformly moving along a circle change?
- 3. What characterizes the impulse of force?
- 4. Do the internal forces of the system affect its momentum? On the movement of its center of mass?
- 5. How do the couples of forces applied to it affect the motion of the center of mass of the system?
- 6. Under what conditions is the center of mass of the system at rest? moving evenly and in a straight line?
7. In a stationary boat, in the absence of water flow, an adult sits at the stern, and a child sits at the bow of the boat. Which direction will the boat move if they swap places?
In which case the module for moving the boat will be large: 1) if the child moves to the adult at the stern; 2) if an adult goes to a child on the bow of the boat? What will be the displacements of the center of mass of the "boat and two people" system during these movements?