Theorems on the change in momentum of a mechanical system. The principle of possible movements

The system discussed in the theorem can be any mechanical system consisting of any bodies.

Statement of the theorem

The amount of motion (impulse) of a mechanical system is a quantity equal to the sum of the amounts of motion (impulses) of all bodies included in the system. The impulse of external forces acting on the bodies of the system is the sum of the impulses of all external forces acting on the bodies of the system.

( kg m/s)

The theorem on the change in momentum of a system states

The change in the momentum of the system over a certain period of time is equal to the impulse of external forces acting on the system over the same period of time.

Law of conservation of momentum of a system

If the sum of all external forces acting on the system is zero, then the amount of motion (momentum) of the system is a constant quantity.

, we obtain the expression of the theorem on the change in the momentum of the system in differential form:

Having integrated both sides of the resulting equality over an arbitrarily taken period of time between some and , we obtain the expression of the theorem on the change in the momentum of the system in integral form:

Law of conservation of momentum (Law of conservation of momentum) states that the vector sum of the impulses of all bodies of the system is a constant value if the vector sum of external forces acting on the system is equal to zero.

(moment of momentum m 2 kg s −1)

Theorem on the change in angular momentum relative to the center

the time derivative of the moment of momentum (kinetic moment) of a material point relative to any fixed center is equal to the moment of force acting on the point relative to the same center.

dk 0 /dt = M 0 (F ) .

Theorem on the change in angular momentum relative to an axis

the time derivative of the moment of momentum (kinetic moment) of a material point relative to any fixed axis is equal to the moment of the force acting on this point relative to the same axis.

dk x /dt = M x (F ); dk y /dt = M y (F ); dk z /dt = M z (F ) .

Consider a material point M mass m , moving under the influence of force F (Figure 3.1). Let's write down and construct the vector of angular momentum (kinetic momentum) M 0 material point relative to the center O :

Let us differentiate the expression for the angular momentum (kinetic moment k 0) by time:

Because dr /dt = V , then the vector product V ⊗ m ⋅ V (collinear vectors V And m ⋅ V ) is equal to zero. In the same time d(m ⋅ V) /dt = F according to the theorem on the momentum of a material point. Therefore we get that

dk 0 /dt = r ⊗F , (3.3)

Where r ⊗F = M 0 (F ) – vector-moment of force F relative to a fixed center O . Vector k 0 ⊥ plane ( r , m ⊗V ), and the vector M 0 (F ) ⊥ plane ( r ,F ), we finally have

dk 0 /dt = M 0 (F ) . (3.4)

Equation (3.4) expresses the theorem about the change in angular momentum (angular momentum) of a material point relative to the center: the time derivative of the moment of momentum (kinetic moment) of a material point relative to any fixed center is equal to the moment of force acting on the point relative to the same center.

Projecting equality (3.4) onto the axes of Cartesian coordinates, we obtain

dk x /dt = M x (F ); dk y /dt = M y (F ); dk z /dt = M z (F ) . (3.5)

Equalities (3.5) express the theorem about the change in angular momentum (kinetic momentum) of a material point relative to the axis: the time derivative of the moment of momentum (kinetic moment) of a material point relative to any fixed axis is equal to the moment of the force acting on this point relative to the same axis.

Let us consider the consequences following from Theorems (3.4) and (3.5).

Corollary 1. Consider the case when the force F during the entire movement of the point passes through the stationary center O (case of central force), i.e. When M 0 (F ) = 0. Then from Theorem (3.4) it follows that k 0 = const ,

those. in the case of a central force, the angular momentum (kinetic moment) of a material point relative to the center of this force remains constant in magnitude and direction (Figure 3.2).

Figure 3.2

From the condition k 0 = const it follows that the trajectory of a moving point is a flat curve, the plane of which passes through the center of this force.

Corollary 2. Let M z (F ) = 0, i.e. force crosses the axis z or parallel to it. In this case, as can be seen from the third of equations (3.5), k z = const ,

those. if the moment of force acting on a point relative to any fixed axis is always zero, then the angular momentum (kinetic moment) of the point relative to this axis remains constant.

Proof of the theorem on the change in momentum

Let the system consist of material points with masses and accelerations. We divide all forces acting on the bodies of the system into two types:

External forces are forces acting from bodies not included in the system under consideration. The resultant of external forces acting on a material point with number i let's denote

Internal forces are the forces with which the bodies of the system itself interact with each other. The force with which on the point with the number i the point with the number is valid k, we will denote , and the force of influence i th point on k th point - . Obviously, when , then

Using the introduced notation, we write Newton’s second law for each of the material points under consideration in the form

Considering that and summing up all the equations of Newton’s second law, we get:

The expression represents the sum of all internal forces acting in the system. According to Newton’s third law, in this sum, each force corresponds to a force such that, therefore, it holds Since the entire sum consists of such pairs, the sum itself is zero. Thus, we can write

Using the notation for the momentum of the system, we obtain

By introducing into consideration the change in the momentum of external forces , we obtain the expression of the theorem on the change in the momentum of the system in differential form:

Thus, each of the last equations obtained allows us to state: a change in the momentum of the system occurs only as a result of the action of external forces, and internal forces cannot have any influence on this value.

Having integrated both sides of the resulting equality over an arbitrarily taken time interval between some and , we obtain the expression of the theorem on the change in the momentum of the system in integral form:

where and are the values ​​of the amount of motion of the system at moments of time and, respectively, and is the impulse of external forces over a period of time. In accordance with what was said earlier and the introduced notations,

In the same way as for one material point, we will derive a theorem on the change in momentum for the system in various forms.

Let's transform the equation (theorem on the movement of the center of mass of a mechanical system)

in the following way:

;

;

The resulting equation expresses the theorem about the change in the momentum of a mechanical system in differential form: the derivative of the momentum of a mechanical system with respect to time is equal to the main vector of external forces acting on the system .

In projections onto Cartesian coordinate axes:

; ; .

Taking the integrals of both sides of the last equations over time, we obtain a theorem about the change in the momentum of a mechanical system in integral form: the change in the momentum of a mechanical system is equal to the momentum of the main vector of external forces acting on the system .

.

Or in projections onto Cartesian coordinate axes:

; ; .

Corollaries from the theorem (laws of conservation of momentum)

The law of conservation of momentum is obtained as special cases of the theorem on the change in momentum for a system depending on the characteristics of the system of external forces. Internal forces can be any, since they do not affect changes in momentum.

There are two possible cases:

1. If the vector sum of all external forces applied to the system is equal to zero, then the amount of motion of the system is constant in magnitude and direction

2. If the projection of the main vector of external forces onto any coordinate axis and/or and/or is equal to zero, then the projection of the momentum on these same axes is a constant value, i.e. and/or and/or respectively.

Similar entries can be made for a material point and for a material point.

The task. From a gun whose mass M, a projectile of mass flies out in a horizontal direction m with speed v. Find speed V guns after firing.

Solution. All external forces acting on the mechanical weapon-projectile system are vertical. This means, based on the corollary to the theorem on the change in the momentum of the system, we have: .

The amount of movement of the mechanical system before firing:

The amount of movement of the mechanical system after the shot:

.

Equating the right-hand sides of the expressions, we obtain that

.

The “-” sign in the resulting formula indicates that after firing the gun will roll back in the direction opposite to the axis Ox.

EXAMPLE 2. A stream of liquid with density flows at a speed V from a pipe with cross-sectional area F and hits a vertical wall at an angle. Determine the fluid pressure on the wall.

SOLUTION. Let us apply the theorem on the change in momentum in integral form to a volume of liquid with a mass m hitting a wall over a period of time t.

MESHCHERSKY EQUATION

(basic equation of the dynamics of a body of variable mass)

In modern technology, cases arise when the mass of a point and a system does not remain constant during movement, but changes. So, for example, during the flight of space rockets, due to the ejection of combustion products and individual unnecessary parts of the rockets, the change in mass reaches 90-95% of the total initial value. But not only space technology can be an example of the dynamics of variable mass motion. In the textile industry, there are significant changes in the mass of various spindles, bobbins, and rolls at modern operating speeds of machines and machines.

Let us consider the main features associated with changes in mass, using the example of the translational motion of a body of variable mass. The basic law of dynamics cannot be directly applied to a body of variable mass. Therefore, we obtain differential equations of motion of a point of variable mass, applying the theorem on the change in the momentum of the system.

Let the point have mass m+dm moves at speed. Then a certain particle with a mass is separated from the point dm moving at speed.

The amount of motion of the body before the particle comes off:

The amount of motion of a system consisting of a body and a detached particle after its separation:

Then the change in momentum:

Based on the theorem about the change in momentum of the system:

Let us denote the quantity - the relative velocity of the particle:

Let's denote

Size R called reactive force. Reactive force is the engine thrust caused by the ejection of gas from the nozzle.

Finally we get

-

This formula expresses the basic equation of the dynamics of a body of variable mass (Meshchersky formula). From the last formula it follows that the differential equations of motion of a point of variable mass have the same form as for a point of constant mass, except for the additional reactive force applied to the point due to the change in mass.

The basic equation for the dynamics of a body of variable mass indicates that the acceleration of this body is formed not only due to external forces, but also due to the reactive force.

Reactive force is a force similar to that felt by the person shooting - when shooting from a pistol, it is felt by the hand; When shooting from a rifle, it is perceived by the shoulder.

Tsiolkovsky's first formula (for a single-stage rocket)

Let a point of variable mass or a rocket move in a straight line under the influence of only one reactive force. Since for many modern jet engines, where is the maximum reactive force (engine thrust) allowed by the engine design; - the force of gravity acting on the engine located on the earth's surface. Those. the above allows us to neglect the component in the Meshchersky equation and accept this equation in the form for further analysis: ,

Let's denote:

Fuel reserve (for liquid jet engines - the dry mass of the rocket (its remaining mass after burning out all the fuel);

The mass of particles separated from the rocket; is considered as a variable value, varying from to .

Let us write the equation of rectilinear motion of a point of variable mass in the following form:

Since the formula for determining the variable mass of a rocket is

Therefore, the equations of motion of a point Taking the integrals of both sides we get

Where - characteristic speed- this is the speed that a rocket acquires under the influence of thrust after all particles have been erupted from the rocket (for liquid jet engines - after all the fuel has burned out).

Placed outside the integral sign (which can be done on the basis of the mean value theorem known from higher mathematics) is the average speed of particles ejected from the rocket.

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Quantity of movement

Momentum of a material point - a vector quantity equal to the product of the mass of a point and its velocity vector.

The unit of measurement for momentum is (kg m/s).

Mechanical system momentum - a vector quantity equal to the geometric sum (principal vector) of the momentum of a mechanical system is equal to the product of the mass of the entire system and the speed of its center of mass.

When a body (or system) moves so that its center of mass is stationary, then the amount of motion of the body is equal to zero (for example, rotation of the body around a fixed axis passing through the center of mass of the body).

In the case of complex motion, the amount of motion of the system will not characterize the rotational part of the motion when rotating around the center of mass. That is, the amount of motion characterizes only the translational motion of the system (together with the center of mass).

Impulse force

The impulse of a force characterizes the action of a force over a certain period of time.

Force impulse over a finite period of time is defined as the integral sum of the corresponding elementary impulses.

Theorem on the change in momentum of a material point

(in differential forms e ):

The time derivative of the momentum of a material point is equal to the geometric sum of the forces acting on the points.

(V integral form ):

The change in the momentum of a material point over a certain period of time is equal to the geometric sum of the impulses of forces applied to the point during this period of time.

Theorem on the change in momentum of a mechanical system

(in differential form ):

The time derivative of the momentum of the system is equal to the geometric sum of all external forces acting on the system.

(in integral form ):

The change in the momentum of a system over a certain period of time is equal to the geometric sum of the impulses of external forces acting on the system during this period of time.

The theorem allows one to exclude obviously unknown internal forces from consideration.

The theorem on the change in momentum of a mechanical system and the theorem on the motion of the center of mass are two different forms of the same theorem.

Law of conservation of momentum of a system

1. If the sum of all external forces acting on the system is equal to zero, then the vector of the system's momentum will be constant in direction and magnitude.
2. If the sum of the projections of all acting external forces onto any arbitrary axis is equal to zero, then the projection of the momentum onto this axis is a constant value.

conclusions:

1. Conservation laws indicate that internal forces cannot change the total amount of motion of the system.
2. The theorem on the change in momentum of a mechanical system does not characterize the rotational motion of a mechanical system, but only the translational one.

An example is given: Determine the momentum of a disk of a certain mass if its angular velocity and size are known.

Calculation example of a spur gear
An example of calculating a spur gear. The choice of material, calculation of permissible stresses, calculation of contact and bending strength have been carried out.

An example of solving a beam bending problem
In the example, diagrams of transverse forces and bending moments were constructed, a dangerous section was found and an I-beam was selected. The problem analyzed the construction of diagrams using differential dependencies and carried out a comparative analysis of various cross sections of the beam.

An example of solving a shaft torsion problem
The task is to test the strength of a steel shaft at a given diameter, material and allowable stress. During the solution, diagrams of torques, shear stresses and twist angles are constructed. The shaft's own weight is not taken into account

An example of solving a problem of tension-compression of a rod
The task is to test the strength of a steel bar at specified permissible stresses. During the solution, diagrams of longitudinal forces, normal stresses and displacements are constructed. The rod's own weight is not taken into account

Application of the theorem on conservation of kinetic energy
An example of solving a problem using the theorem on the conservation of kinetic energy of a mechanical system

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An example of solving a problem to determine the speed and acceleration of a point using given equations of motion

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An example of solving a problem to determine the velocities and accelerations of points of a rigid body during plane-parallel motion

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An example of solving the problem of determining the forces in the rods of a flat truss using the Ritter method and the method of cutting nodes

Application of the theorem on the change in angular momentum
An example of solving a problem using the theorem on the change in kinetic momentum to determine the angular velocity of a body rotating around a fixed axis.

(Fragments of a mathematical symphony)

The connection between the impulse of force and the basic equation of Newtonian dynamics is expressed by the theorem on the change in the momentum of a material point.

Theorem. The change in the momentum of a material point over a certain period of time is equal to the impulse of the force () acting on the material point over the same period of time. The mathematical proof of this theorem can be called a fragment of a mathematical symphony. Here he is.

The differential momentum of a material point is equal to the elementary impulse of the force acting on the material point. Integrating expression (128) for the differential momentum of a material point, we have

(129)

The theorem has been proven and mathematicians consider their mission completed, but engineers, whose destiny is to sacredly believe in mathematicians, have questions when using the proven equation (129). But they are firmly blocked by the sequence and beauty of mathematical operations (128 and 129), which fascinate and encourage us to call them a fragment of a mathematical symphony. How many generations of engineers agreed with mathematicians and were in awe of the mystery of their mathematical symbols! But then there was an engineer who disagreed with the mathematicians and asked them questions.

Dear mathematicians! Why doesn’t any of your textbooks on theoretical mechanics discuss the process of applying your symphonic result (129) in practice, for example, when describing the process of accelerating a car? The left side of equation (129) is very clear. The car starts acceleration from speed and ends it, for example, at speed. It is quite natural that equation (129) becomes

And the first question immediately arises: how can we determine from equation (130) the force under the influence of which the car is accelerated to a speed of 10 m/s? The answer to this question is not found in any of the countless textbooks on theoretical mechanics. Let's go further. After acceleration, the car begins to move uniformly at a speed of 10 m/s. What force moves the car?????????? I have no choice but to blush along with the mathematicians. The first law of Newtonian dynamics states that when a car moves uniformly, no forces act on it, and the car, figuratively speaking, sneezes at this law, consumes gasoline and does work, moving, for example, a distance of 100 km. Where is the force that did the work to move the car 100 km? The symphonic mathematical equation (130) is silent, but life goes on and demands an answer. We start looking for him.

Since the car moves rectilinearly and uniformly, the force moving it is constant in magnitude and direction and equation (130) becomes

(131)

So, equation (131) in this case describes the accelerated motion of the body. What is the force equal to? How to express its change over time? Mathematicians prefer to bypass this question and leave it to engineers, believing that they must search for the answer to this question. Engineers have only one option left - to take into account that if, after the completion of the accelerated motion of the body, a phase of uniform motion begins, which is accompanied by the action of a constant force, present equation (131) for the moment of transition from accelerated to uniform motion in this form

(132)

The arrow in this equation does not mean the result of integrating this equation, but the process of transition from its integral form to a simplified form. The force in this equation is equivalent to the average force that changed the momentum of the body from zero to a final value. So, dear mathematicians and theoretical physicists, the absence of your method for determining the magnitude of your impulse forces us to simplify the procedure for determining force, and the absence of a method for determining the time of action of this force generally puts us in a hopeless position and we are forced to use an expression to analyze the process of changing the momentum of a body . The result is that the longer the force acts, the greater its impulse. This clearly contradicts the long-established idea that the shorter the duration of its action, the greater the force impulse.

Let us pay attention to the fact that the change in the momentum of a material point (impulse of force) during its accelerated movement occurs under the action of Newtonian force and forces of resistance to movement, in the form of forces formed by mechanical resistances and the force of inertia. But Newtonian dynamics in the vast majority of problems ignores the force of inertia, and Mechanodynamics states that a change in the momentum of a body during its accelerated motion occurs due to the excess of the Newtonian force over the forces of resistance to movement, including the force of inertia.

When a body moves in slow motion, for example, a car with the gear turned off, there is no Newtonian force, and the change in the momentum of the car occurs due to the excess of the motion resistance forces over the inertia force that moves the car when it moves slowly.

How can we now return the results of the noted “symphonic” mathematical actions (128) to the mainstream of cause-and-effect relationships? There is only one way out - to find a new definition of the concepts “impulse of force” and “impact force”. To do this, divide both sides of equation (132) by time t. As a result we will have

. (133)

Let us note that the expression mV/t is the rate of change of momentum (mV/t) of a material point or body. If we take into account that V/t is acceleration, then mV/t is the force that changes the momentum of the body. The same dimension on the left and right of the equal sign gives us the right to call the force F a shock force and denote it by the symbol, and the impulse S - a shock impulse and denote it by the symbol. This leads to a new definition of impact force. The impact force acting on a material point or body is equal to the ratio of the change in the momentum of the material point or body to the time of this change.

Let us pay special attention to the fact that only the Newtonian force participates in the formation of the shock impulse (134), which changed the speed of the car from zero to the maximum - , therefore equation (134) belongs entirely to Newtonian dynamics. Since it is much easier to determine the magnitude of velocity experimentally than it is to determine acceleration, formula (134) is very convenient for calculations.

This unusual result follows from equation (134).

Let us pay attention to the fact that according to the new laws of mechanodynamics, the generator of the force impulse during the accelerated movement of a material point or body is Newtonian force. It forms the acceleration of the movement of a point or body, at which an inertial force automatically arises, directed opposite to the Newtonian force and the impact Newtonian force must overcome the action of the inertial force, therefore the inertial force must be represented in the balance of forces on the left side of equation (134). Since the inertial force is equal to the mass of the point or body multiplied by the deceleration that it forms, then equation (134) becomes

(136)

Dear mathematicians! You see what form the mathematical model has taken on, describing the shock impulse, which accelerates the movement of the impacted body from zero speed to maximum V (11). Now let’s check its operation in determining the impact impulse, which is equal to the impact force that fired the 2nd power unit of the SShG (Fig. 120), and we will leave you with your useless equation (132). In order not to complicate the presentation, we will leave formula (134) alone for now and use formulas that give average values ​​of forces. You see in what position you put an engineer trying to solve a specific problem.

Let's start with Newtonian dynamics. Experts found that the 2nd power unit rose to a height of 14 m. Since it rose in the field of gravity, at a height of h = 14 m its potential energy turned out to be equal to

and the average kinetic energy was equal to

Rice. 120. Photo of the turbine room before the disaster

From the equality of kinetic (138) and potential (137) energies, the average rate of rise of the power unit follows (Fig. 121, 122)

Rice. 121. Photon of the turbine room after the disaster

According to the new laws of mechanodynamics, the rise of the power unit consisted of two phases (Fig. 123): the first phase OA - accelerated rise and the second phase AB - slow rise , , .

The time and distance of their action are approximately equal (). Then the kinematic equation of the accelerated phase of raising the power unit will be written as follows:

. (140)

Rice. 122. View of the well of the power unit and the power unit itself after the disaster

The law of change in the rate of rise of the power unit in the first phase has the form

. (141)

Rice. 123. Regularity of changes in flight speed V of a power unit

Substituting time from equation (140) into equation (141), we have

. (142)

The block lifting time in the first phase is determined from formula (140)

. (143)

Then the total time for raising the power unit to a height of 14 m will be equal to . The mass of the power unit and cover is 2580 tons. According to Newtonian dynamics, the force that lifted the power unit is equal to

Dear mathematicians! We follow your symphonic mathematical results and write down your formula (129), following from Newtonian dynamics, to determine the shock pulse that fired the 2nd power unit

and ask a basic question: how to determine the duration of the shock pulse that fired the 2nd power unit????????????

Dear!!! Remember how much chalk was written on the blackboards by generations of your colleagues, abstrusely teaching students how to determine the shock impulse, and no one explained how to determine the duration of the shock impulse in each specific case. You will say that the duration of the shock pulse is equal to the time interval of the change in the speed of the power unit from zero to, we will assume, the maximum value of 16.75 m/s (139). It is in formula (143) and is equal to 0.84 s. We agree with you for now and determine the average value of the shock impulse

The question immediately arises: why is the magnitude of the shock impulse (146) less than the Newtonian force of 50600 tons? You, dear mathematicians, have no answer. Let's go further.

According to Newtonian dynamics, the main force that resisted the rise of the power unit was gravity. Since this force is directed against the movement of the power unit, it generates a deceleration that is equal to the acceleration of free fall. Then the gravitational force acting on the power unit flying upward is equal to

Newton's dynamics does not take into account other forces that prevented the action of the Newtonian force of 50,600 tons (144), and mechanodynamics states that the rise of the power unit was also resisted by an inertial force equal to

The question immediately arises: how to find the amount of deceleration in the movement of the power unit? Newtonian dynamics is silent, but mechanodynamics answers: at the moment of action of the Newtonian force, which lifted the power unit, it was resisted by: the force of gravity and the force of inertia, therefore the equation of the forces acting on the power unit at that moment is written as follows.

The amount of motion is a measure of mechanical movement, if mechanical movement turns into mechanical. For example, the mechanical movement of a billiard ball (Fig. 22) before impact turns into mechanical movement of the balls after impact. For a point, the momentum is equal to the product .

The measure of the force in this case is the force impulse

. (9.1)

Momentum determines the action of force over a period of time . For a material point, the theorem on the change in momentum can be used in differential form
(9.2) or integral (finite) form
. (9.3)

The change in the momentum of a material point over a certain period of time is equal to the impulse of all forces applied to the point during the same time.

Figure 22

When solving problems, Theorem (9.3) is more often used in projections onto coordinate axes
;

; (9.4)

.

Using the theorem on the change in the momentum of a point, it is possible to solve problems in which a point or body moving translationally is acted upon by constant or variable forces that depend on time, and the given and sought quantities include the time of movement and velocities at the beginning and end of the movement. Problems using the theorem are solved in the following sequence:

1. choose a coordinate system;

2. depict all given (active) forces and reactions acting on a point;

3. write down a theorem about the change in the momentum of a point in projections onto the selected coordinate axes;

4. determine the required quantities.

EXAMPLE 12.

A hammer weighing G=2t falls from a height h=1m onto the workpiece in time t=0.01s and stamps the part (Fig. 23). Determine the average pressure force of the hammer on the workpiece.

The change in the momentum of a mechanical system over a certain period of time is equal to the sum of the impulses of external forces acting during the same time. Sometimes it is more convenient to use the theorem on the change in momentum in projection on the coordinate axes
; (9.8)
. (9.9)

The law of conservation of momentum states that in the absence of external forces, the momentum of a mechanical system remains constant. The action of internal forces cannot change the momentum of the system. From equation (9.6) it is clear that when
,
.

If
, That
or
.

D

propeller or propeller, jet propulsion. Squids move in jerks, throwing water out of the muscular sac like a water cannon (Fig. 25). The repelled water has a certain amount of motion directed backwards. The squid receives the corresponding speed forward movement due to reactive traction force , since before the squid jumps out the force balanced by gravity .

Application of the theorem on the change in momentum allows us to exclude all internal forces from consideration.

EXAMPLE 13.

A winch A with a drum of radius r is installed on a railway platform free-standing on the rails (Fig. 26). The winch is designed to move a load B with a mass m 1 along the platform. Weight of the platform with winch m 2. The winch drum rotates according to the law
. At the initial moment of time the system was mobile. Neglecting friction, find the law of change in the speed of the platform after turning on the winch.

R SOLUTION.

1. Consider the platform, winch and load as a single mechanical system, which is acted upon by external forces: gravity of the load and platforms and reactions And
.

2. Since all external forces are perpendicular to the x axis, i.e.
, we apply the law of conservation of momentum of a mechanical system in projection onto the x-axis:
. At the initial moment of time the system was motionless, therefore,

Let us express the amount of motion of the system at an arbitrary moment in time. The platform moves forward at a speed , the load undergoes a complex movement consisting of relative movement along the platform with a speed and portable movement together with the platform at speed ., where
. The platform will move in the direction opposite to the relative movement of the load.

EXAMPLE 14.

Where ,
-- the amount of motion of the plate and load, respectively.

;
, Where --absolute speed of the load D. From equality (1) it follows that K 1x + K 2x =C 1 or m 1 u x +m 2 V Dx =C 1. (2) To determine V Dx, consider the motion of the load D as complex, considering its motion relative to the plate relative, and the motion of the plate itself portable, then
, (3)
;or in projection onto the x axis: . (4) Let's substitute (4) into (2):
. (5) We determine the integration constant C 1 from the initial conditions: at t=0 u=u 0 ; (m 1 +m 2)u 0 =C 1. (6) Substituting the value of the constant C 1 into equation (5), we obtain

m/s.