Theorem on the change in momentum of a point

Since the mass of a point is constant, and its acceleration, the equation expressing the basic law of dynamics can be represented as

The equation simultaneously expresses the theorem on the change in the momentum of a point in differential form: time derivative of the momentum of a point is equal to the geometric sum of the forces acting on the point.

Let's integrate this equation. Let the mass point m, moving under the action of a force (Fig. 15), has at the moment t\u003d 0 speed, and at the moment t 1 - speed.

Fig.15

Let us then multiply both sides of the equality by and take definite integrals from them. In this case, on the right, where the integration is over time, the limits of the integrals will be 0 and t 1 , and on the left, where the speed is integrated, the limits of the integral will be the corresponding values ​​of the speed and . Since the integral of equals , then as a result we get:

.

The integrals on the right are the impulses of the acting forces. So we end up with:

.

The equation expresses the theorem on the change in the momentum of a point in the final form: the change in the momentum of a point over a certain period of time is equal to the geometric sum of the impulses of all forces acting on the point over the same period of time ( rice. 15).

When solving problems, instead of a vector equation, equations in projections are often used.

In the case of rectilinear motion along the axis Oh the theorem is expressed by the first of these equations.

Example 9 Find the law of motion of a material point of mass m moving along the axis X under the action of a force constant in modulus F(Fig. 16) under initial conditions: , at .

Fig.16

Decision. Let us compose a differential equation of motion of a point in the projection onto the axis X: . Integrating this equation, we find: . The constant is determined from the initial condition for the velocity and is equal to . Finally

.

Further, taking into account that v = dx/dt, we arrive at the differential equation: , integrating which we get

The constant is determined from the initial condition for the coordinate of the point. She is equal. Therefore, the law of motion of a point has the form

Example 10. Load weight R(Fig. 17) begins to move from rest along a smooth horizontal plane under the action of a force F=kt. Find the law of motion of the load.

Fig.17

Decision. We choose the origin of the coordinate system ABOUT in the initial position of the load and direct the axle X in the direction of movement (Fig. 17). Then the initial conditions look like: x(t = 0) = 0,v( t = 0) = 0. Forces act on the load F,P and the reaction force of the plane N. The projections of these forces on the axis X matter Fx = F = kt, Rx = 0, N x= 0, so the corresponding equation of motion can be written as follows: . Separating the variables in this differential equation and then integrating, we get: v = gkt 2 /2P + C one . Substituting the initial data ( v(0) = 0), we find that C 1 = 0, and we get the law of speed change .

The last expression, in turn, is a differential equation, integrating which we will find the law of motion of a material point: . The constant entering here is determined from the second initial condition X(0) = 0. It is easy to see that . Finally

Example 11. On a load at rest on a horizontal smooth plane (see Fig. 17) at a distance a from the origin, begins to act in the positive direction of the axis x force F=k 2 (P/g)x, where R - cargo weight. Find the law of motion of the load.

Decision. The equation of motion of the considered load (material point) in the projection onto the axis X

The initial conditions of equation (1) have the form: x(t = 0) = a, v( t = 0) = 0.

We represent the time derivative of velocity entering equation (1) as follows:

.

Substituting this expression into equation (1) and reducing by ( P/g), we get

Separating the variables in the last equation, we find that . Integrating the latter, we have: . Using initial conditions , we get , and, therefore,

, . (2)

Since the force acts on the load in the positive direction of the axis X, it is clear that it must also move in the same direction. Therefore, in solution (2), the plus sign should be chosen. Replacing further in the second expression (2) with , we obtain a differential equation for determining the law of movement of the load. Whence, separating the variables, we have

.

Integrating the latter, we find: . After finding the constant, we finally get

Example 12. Ball M masses m(Fig.18) falls without initial velocity under the action of gravity. As the ball falls, it experiences resistance , where – constant drag coefficient. Find the law of motion of the ball.

Fig.18

Decision. Let us introduce a coordinate system with the origin at the point where the ball is located at t = 0, directing the axis at vertically down (Fig. 18). The differential equation of motion of the ball in the projection onto the axis at then has the form

The initial conditions for the ball are written as follows: y(t = 0) = 0, v( t = 0) = 0.

Separating variables in equation (1)

and integrating, we find: , where . Or after finding a constant

or . (2)

It follows from this that the limiting speed, i.e. speed at , is equal to .

To find the law of motion, we replace v in equation (2) by dy/dt. Then, integrating the resulting equation with allowance for the initial condition, we finally find

.

Example 13 Research submarine of spherical shape and mass m= = 1.5×10 5 kg starts to sink with the engines off, having a horizontal speed v X 0 = 30 m/s and negative buoyancy R 1 = 0.01mg, where is the vector sum of the Archimedean buoyancy force Q and gravity mg acting on the boat (Fig. 20). Water resistance force , kg/s. Determine the equations of motion of the boat and its trajectory.

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

Statement of the theorem

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

( kg m/s)

The theorem on the change in the momentum of the system states

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

Law of conservation of momentum of the system

If the sum of all external forces acting on the system is equal to zero, then the momentum (momentum) of the system is a constant value.

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

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

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

(Moment of momentum m 2 kg s −1)

Theorem on the change in the angular momentum about the center

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

dk 0 /dt = M 0 (F ) .

Theorem on the change in the angular momentum about the axis

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

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

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

Differentiate the expression for moment of momentum (kinetic moment k 0) by time:

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

dk 0 /dt = r ⊗F , (3.3)

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

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

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

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

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

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

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

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

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

Figure 3.2

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

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

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

Proof of the momentum change theorem

Let the system consist of material points with masses and accelerations . All forces acting on the bodies of the system can be divided into two types:

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

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

Using the introduced notation, we write Newton's second law for each of the considered material points in the form

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

The expression is the sum of all internal forces acting in the system. According to Newton's third law, in this sum, each force corresponds to a force such that and, therefore, is fulfilled Since the whole sum consists of such pairs, the sum itself is equal to zero. Thus, one can write

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

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

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

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

where and are the values ​​of the amount of motion of the system at the moments of time and, respectively, and is the impulse of external forces over a period of time . In accordance with the above and the introduced notation,

Differential equation of motion of a material point under the action of a force F can be represented in the following vector form:

Since the mass of a point m is assumed to be constant, then it can be introduced under the sign of the derivative. Then

Formula (1) expresses the theorem on the change in the momentum of a point in differential form: the first time derivative of the momentum of a point is equal to the force acting on the point.

In projections onto the coordinate axes (1) can be represented as

If both sides of (1) are multiplied by dt, then we obtain another form of the same theorem - the momentum theorem in differential form:

those. the differential of the momentum of a point is equal to the elementary impulse of the force acting on the point.

Projecting both parts of (2) onto the coordinate axes, we obtain

Integrating both parts of (2) from zero to t (Fig. 1), we have

where is the speed of the point at the moment t; - speed at t = 0;

S- momentum of force over time t.

The expression in the form (3) is often called the momentum theorem in finite (or integral) form: the change in the momentum of a point over any period of time is equal to the momentum of the force over the same period of time.

In projections onto the coordinate axes, this theorem can be represented in the following form:

For a material point, the theorem on the change in momentum in any of the forms, in essence, does not differ from the differential equations of motion of a point.

Theorem on the change in the momentum of the system

The amount of motion of the system will be called the vector quantity Q, equal to the geometric sum (principal vector) of the momentum of all points of the system.

Consider a system consisting of n material points. Let us compose differential equations of motion for this system and add them term by term. Then we get:

The last sum by the property of internal forces is equal to zero. Besides,

We finally find:

Equation (4) expresses the theorem on the change in the momentum of the system in differential form: the time derivative of the momentum of the system is equal to the geometric sum of all external forces acting on the system.

Let's find another expression of the theorem. Let at the moment t= 0 the momentum of the system is Q0, and at the moment of time t1 becomes equal Q1. Then, multiplying both sides of equality (4) by dt and integrating, we get:

Or where:

(S- force impulse)

since the integrals on the right give the impulses of external forces,

equation (5) expresses the theorem on the change in the momentum of the system in integral form: the change in the amount of motion of the system over a certain period of time is equal to the sum of the impulses of external forces acting on the system over the same period of time.

In projections on the coordinate axes, we will have:

Law of conservation of momentum

From the theorem on the change in the momentum of the system, the following important consequences can be obtained:

1. Let the sum of all external forces acting on the system be equal to zero:

Then it follows from Eq. (4) that, in this case, Q=const.

Thus, if the sum of all external forces acting on the system is equal to zero, then the momentum vector of the system will be constant in modulus and direction.

2. Let the external forces acting on the system be such that the sum of their projections on some axis (for example, Ox) is equal to zero:

Then it follows from equations (4`) that in this case Q = const.

Thus, if the sum of the projections of all acting external forces on some axis is equal to zero, then the projection of the momentum of the system on this axis is a constant value.

These results express law of conservation of momentum of the system. It follows from them that internal forces cannot change the total momentum of the system.

Let's look at some examples:

· P h e n i e o f recoil or recoil. If we consider a rifle and a bullet as one system, then the pressure of the powder gases when fired will be an internal force. This force cannot change the total momentum of the system. But since the propellant gases, acting on the bullet, give it a certain amount of movement directed forward, they must simultaneously tell the rifle the same amount of movement in the opposite direction. This will cause the rifle to move backward, i.e. so-called return. A similar phenomenon occurs when firing from a gun (rollback).

· Operation of the propeller (propeller). The propeller informs a certain mass of air (or water) of motion along the axis of the propeller, throwing this mass back. If we consider the ejected mass and the aircraft (or ship) as one system, then the interaction forces of the propeller and the medium as internal cannot change the total momentum of this system. Therefore, when a mass of air (water) is thrown back, the aircraft (or vessel) obtains the corresponding forward speed, such that the total momentum of the system under consideration remains equal to zero, since it was zero before the start of movement.

A similar effect is achieved by the action of oars or paddle wheels.

· Rocket propulsion. In a rocket projectile (rocket), gaseous products of fuel combustion are ejected at high speed from a hole in the tail of the rocket (from the nozzle of a jet engine). The pressure forces acting in this case will be internal forces and they cannot change the total momentum of the rocket-powder gases system. But since the escaping gases have a certain amount of motion directed backwards, the rocket receives in this case the corresponding forward speed.

Theorem of moments about the axis.

Consider a material point of mass m moving under the influence of a force F. Let us find for it the dependence between the moment of the vectors mV And F about some fixed Z-axis.

m z (F) = xF - yF (7)

Similarly for the quantity m (mV), if taken out m bracket will be

m z (mV) \u003d m (xV - yV)(7`)

Taking time derivatives of both sides of this equality, we find

On the right side of the resulting expression, the first parenthesis is 0, since dx/dt=V and dу/dt=V, while the second bracket according to formula (7) is equal to

m z (F), since according to the basic law of dynamics:

Finally we will have (8)

The resulting equation expresses the theorem of moments about the axis: the time derivative of the angular momentum of a point about some axis is equal to the moment of the acting force about the same axis. A similar theorem also holds for moments about any center O.

(Fragments of a mathematical symphony)

The connection of the force impulse with 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 amount of motion of a material point for a certain period of time is equal to the impulse of the force () acting on the material point for the same period of time. The mathematical proof of this theorem can be called a fragment of a mathematical symphony. There he is.

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

(129)

The theorem is proved and mathematicians consider their mission completed, and engineers, whose fate is to sacredly believe mathematicians, have questions when using the proven equation (129). But they are firmly blocked by the sequence and beauty of mathematical actions (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 trembled at the mystery of their mathematical symbols! But then there was an engineer who disagreed with the mathematicians and asked them questions.

Dear mathematicians! Why is it that none of your textbooks on theoretical mechanics discusses 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 extremely clear. The car starts acceleration from a speed and finishes it, for example, at a speed of . It is quite natural that equation (129) becomes

And the first question immediately arises: how can we determine the force from equation (130), under the influence of which the car is accelerated to a speed of 10 m/s? There is no answer to this question in any of the innumerable textbooks on theoretical mechanics. Let's go further. After acceleration, the car begins to move uniformly with the achieved speed of 10m/s. What is the force that drives 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 on this law, consumes gasoline and does work, moving, for example, a distance of 100 km. And where is the force that has done the work to move the car 100 km? The symphonic mathematical equation (130) is silent, but life goes on and requires an answer. We start looking for it.

Since the car is moving in a straight line and uniformly, the force that moves 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 sidestep this question and leave it to the engineers, believing that they should look for the answer to this question. Engineers have one possibility 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 a constant force, represent 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 the final value . So, dear mathematicians and theoretical physicists, the absence of your method for determining the magnitude of your momentum forces us to simplify the procedure for determining the force, and the lack of a method for determining the duration of this force generally puts us in a hopeless situation and we are forced to use the expression to analyze the process of changing the momentum of the body . As a result, the longer the force acts, the greater its momentum. This clearly contradicts the long-established ideas that the impulse of force is greater, the shorter the time of its action.

Let us pay attention to the fact that the change in the momentum of a material point (force impulse) during its accelerated movement occurs under the action of the Newtonian force and the 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 asserts that the change in the momentum of a body during its accelerated movement 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 a gear off, there is no Newtonian force, and the change in the momentum of the car occurs due to the excess of the forces of resistance to movement over the force of inertia that moves the car during its slow motion.

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

. (133)

Let's pay attention to the fact that the expression mV / t is the rate of change in the momentum (mV / t) of a material point or body. If we take into account that V / t is acceleration, then mV / t is a 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 the impact force and designate it with the symbol , and the impulse S - the impact impulse and designate it with the symbol . From this follows a new definition of impact force. 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 is involved in the formation of the shock impulse (134), which changed the speed of the car from zero to the maximum value - , therefore, equation (134) belongs entirely to Newtonian dynamics. Since it is much easier to fix the speed value experimentally than accelerations, formula (134) is very convenient for calculations.

Equation (134) implies such an unusual result.

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 the Newtonian force. It generates an acceleration of the movement of a point or body, at which an inertia force automatically arises, directed opposite to the Newtonian force, and the impact Newtonian force must overcome the action of the inertia force, therefore the inertia force must be represented in the balance of forces on the left side of equation (134). Since the force of inertia is equal to the mass of a point or body, multiplied by the deceleration , which it forms, then equation (134) becomes

(136)

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

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

and the average kinetic energy was

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

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

Rice. 121. Photon of the machine 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 - an accelerated rise and the second phase AB - a slow rise , , .

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

. (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 lifting speed of the power unit in the first phase has the form

. (141)

Rice. 123. The pattern of change in the speed V of the flight of the power unit

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

. (142)

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

. (143)

Then the total time of lifting the power unit to a height of 14m will be equal to . The mass of the power unit and the cover is 2580 tons. According to Newton's 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), which follows from Newton's dynamics, to determine the shock impulse that fired the 2nd power unit

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

Dear!!! Remember how much chalk the generations of your colleagues wrote on the educational boards, abstrusely teaching students how to determine the impact impulse and no one explained how to determine the duration of the impact impulse in each specific case. You will say the duration of the shock impulse is equal to the time interval for changing the speed of the power unit from zero to, we 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 value of the shock impulse (146) less than the Newtonian force of 50600 tons? The answer, you, dear mathematicians, no. Let's go further.

According to Newton's dynamics, the main force that resisted the lifting of the power unit is gravity. Since this force is directed against the movement of the power unit, it generates a deceleration, which is equal to the free fall acceleration. Then the gravitational force acting on the power unit flying upwards is equal to

The dynamics of Newton does not take into account other forces that prevented the action of the Newtonian force of 50600 tons (144), and mechanodynamics claims that the inertia force equal to

The question immediately arises: how to find the magnitude of the deceleration of the movement of the power unit? Newton's dynamics is silent, and mechanodynamics answers: at the moment of action of the Newtonian force that lifted the power unit, it was resisted: gravity and inertia, so the equation of forces acting on the power unit at that moment is written as follows.

Consider a system consisting of material points. Let us compose differential equations of motion (13) for this system and add them term by term. Then we get

The last sum by the property of internal forces is equal to zero. Besides,

Finally we find

Equation (20) expresses the theorem on the change in the momentum of the system in differential form: the time derivative of the momentum of the system is equal to the geometric sum of all external forces acting on the system. In projections onto the coordinate axes it will be:

Let us find another expression of the theorem. Let at the moment of time the momentum of the system is equal to and at the moment it becomes equal to . Then, multiplying both sides of equality (20) by and integrating, we obtain

since the integrals on the right give the impulses of external forces.

Equation (21) expresses the theorem on the change in the momentum of the system in integral form: the change in the momentum of the system over a certain period of time is equal to the sum of the impulses acting on the system of external forces over the same period of time.

In projections onto the coordinate axes it will be:

Let us point out the connection between the proved theorem and the theorem on the motion of the center of mass. Since , then, substituting this value into equality (20) and taking into account that we get , i.e. equation (16).

Therefore, the theorem on the motion of the center of mass and the theorem on the change in the momentum of the system are, in essence, two different forms of the same theorem. In cases where the motion of a rigid body (or a system of bodies) is being studied, any of these forms can equally be used, and equation (16) is usually more convenient to use. For a continuous medium (liquid, gas), when solving problems, they usually use the theorem on the change in the momentum of the system. This theorem also has important applications in the theory of impact (see Ch. XXXI) and in the study of jet propulsion (see § 114).