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Number of movement
Quantity of movement of a material point - a vector quantity equal to the product of the mass of the point and the vector of its velocity.
The unit of momentum is (kg m/s).
Quantity of movement of the mechanical system - vector quantity equal to geometric sum(main 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 in such a way that its center of mass is stationary, then the momentum of the body is zero (for example, the rotation of the body around a fixed axis passing through the center of mass of the body).
In the case of complex motion, the momentum 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 of force
The momentum of a force characterizes the action of a force over a certain period of time.
Impulse of force 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 form e ):
The time derivative of the momentum of a material point is equal to the geometric sum of the forces acting on the points.
(in 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 over this period of time.
Theorem on the change in the 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 amount of motion of the system over a certain period of time is equal to the geometric sum of the impulses of external forces acting on the system over this period of time.
The theorem makes it possible 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 the system
- 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 direction and modulo.
- If the sum of the projections of all acting external forces on any arbitrary axis is equal to zero, then the projection of the momentum on this axis is a constant value.
findings:
- Conservation laws indicate that internal forces cannot change the total momentum of the system.
- The theorem on the change in momentum of a mechanical system does not characterize rotary motion mechanical system, but only progressive.
An example is given: Determine the amount of motion of a disk of a certain mass, if its angular velocity and size are known.
An example of the calculation of a spur gear
An example of the calculation of a spur gear. The choice of material, the calculation of allowable stresses, the calculation of contact and bending strength were carried out.
An example of solving the problem of beam bending
In the example, diagrams of transverse forces and bending moments are plotted, a dangerous section is found, and an I-beam is selected. In the problem, the construction of diagrams using differential dependencies was analyzed, comparative analysis various cross sections beams.
An example of solving the problem of shaft torsion
The task is to test the strength of a steel shaft for a given diameter, material and allowable stresses. During the solution, diagrams of torques, shear stresses and twist angles are built. Self weight of the shaft is not taken into account
An example of solving the problem of tension-compression of a rod
The task is to test the strength of a steel rod at given allowable stresses. Diagrams are built during the solution longitudinal forces, normal stresses and displacements. Self weight of the bar is not taken into account
Application of the kinetic energy conservation theorem
An example of solving the problem of applying the theorem on the conservation of kinetic energy of a mechanical system
Determination of the speed and acceleration of a point according to the given equations of motion
An example of solving the problem of determining the speed and acceleration of a point according to the given equations of motion
Determination of velocities and accelerations of points of a rigid body during plane-parallel motion
An example of solving the problem of determining the velocities and accelerations of points solid body in plane-parallel motion
Determination of Forces in Planar Truss Bars
An example of solving the problem of determining the forces in the bars of a flat truss by the Ritter method and the knot cutting method
Application of the torque change theorem
An example of solving the problem of applying the change theorem angular momentum to determine the angular velocity of a body rotating around a fixed axis.
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 from them definite integrals. 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 is , 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. fifteen).
When solving problems, instead of a vector equation, equations in projections are often used.
When rectilinear motion occurring along the axis Oh the theorem is expressed by the first of these equations.
Questions for self-examination
Formulate the basic laws of mechanics.
What equation is called the basic equation of dynamics?
What is the measure of inertia of solid bodies in translational motion?
Does the weight of the body depend on the location of the body on Earth?
What frame of reference is called inertial?
To what body is the force of inertia of a material point applied and what is its modulus and direction?
Explain the difference between the concepts of "inertia" and "force of inertia"?
To what bodies is the force of inertia applied, how is it directed and by what formula can it be calculated?
What is the principle of kinetostatics?
What are the modules and directions of the tangential and normal forces of inertia of a material point?
What is body weight? What is the SI unit for measuring mass?
What is the measure of inertia of a body?
Write down the basic law of dynamics in vector and differential form?
A constant force acts on a material point. How does the dot move?
What acceleration will the point receive if a force equal to twice the force of gravity acts on it?
After the collision of two material points with masses m 1 =6 kg and m 2 \u003d 24 kg, the first point received an acceleration of 1.6 m / s. What is the acceleration obtained by the second point?
At what motion of a material point is its tangential force of inertia equal to zero and at what is it normal?
What formulas are used to calculate the modules of rotational and centrifugal forces of inertia of a point belonging to a rigid body rotating around a fixed axis?
How is the basic law of point dynamics formulated?
Give the formulation of the law of independence of the action of forces.
write down differential equations motion of a material point in vector and coordinate form.
Formulate the essence of the first and second main tasks of point dynamics.
Give the conditions from which the constants of integration of the differential equations of motion of a material point are determined.
What equations of dynamics are called natural equations of motion of a material point?
What are the two main problems of point dynamics that are solved with the help of differential motions of a material point?
Differential equations of motion of a free material point.
How are the constants determined when integrating the differential equations of motion of a material point?
Determination of the values of arbitrary constants that appear when integrating the differential equations of motion of a material point.
What are the laws free fall body?
What are the laws governing the horizontal and vertical movement of a body thrown at an angle to the horizon in a vacuum? What is the trajectory of its movement and at what angle does the body have the greatest flight range?
How to calculate the impulse of a variable force over a finite period of time?
What is the momentum of a material point?
How to express the elementary work of the force through the elementary path of the point of application of the force, and how - through the increment of the arc coordinate of this point?
On what displacements is the work of gravity: a) positive, b) negative, c) equal to zero?
How to calculate the power of a force applied to a material point rotating around a fixed axis with an angular velocity ?
Formulate a theorem on the change in the momentum of a material point.
Under what conditions does the momentum of a material point not change? Under what conditions does its projection on some axis not change?
Give the formulation of the theorem on the change in the kinetic energy of a material point in differential and finite form.
What is called the moment of momentum of a material point relative to: a) the center, b) the axis?
How is the theorem about the change in the angular momentum of a point relative to the center and relative to the axis formulated?
Under what conditions does the angular momentum of a point about an axis remain unchanged?
How are the moments of momentum of a material point relative to the center and relative to the axis determined? What is the relationship between them?
At what location of the momentum vector of a material point its moment relative to the axis zero?
Why does the trajectory of a material point moving under the action of a central force lie in one plane?
What movement of a point is called rectilinear? Write down the differential equation for the rectilinear motion of a material point.
Write down the differential equations for the plane motion of a material point.
What motion of a material point is described by Lagrange's differential equations of the first kind?
In what cases is a material point called non-free and what are the differential equations of motion of this point?
Give definitions of stationary and nonstationary, holonomic and nonholonomic constraints.
What are two-way relationships? Unilateral?
What is the essence of the principle of release from bonds?
What form do the differential equations of motion of a non-free material point have in the Lagrange form? What is the Lagrange multiplier?
Give the formulation of the dynamical Coriolis theorem.
What is the essence of the Galileo-Newton principle of relativity?
Name the movements in which the Coriolis force of inertia is zero.
What is the magnitude and direction of the translational and Coriolis inertia forces?
What is the difference between the differential equations of relative and absolute motions of a material point?
How are the translational and Coriolis forces of inertia determined in various occasions portable movement?
What is the essence of the principle of relativity of classical mechanics?
What reference systems are called inertial?
What is the condition for the relative rest of a material point?
At what points earth's surface gravity has the greatest and smallest value?
What explains the deviation of falling bodies to the east?
In what direction is a body thrown vertically upwards deflected?
A bucket is lowered into the mine with acceleration a\u003d 4 m / s 2. Tub gravity G=2 kN. Determine the tension in the rope supporting the tub?
Two material points move in a straight line with constant speeds of 10 and 100 m/s. Can it be argued that equivalent systems of forces are applied to these points?
1) it is impossible;
Equal forces are applied to two material points of mass 5 and 15 kg. Compare the numerical values of the acceleration of these points?
1) the accelerations are the same;
2) the acceleration of a point with a mass of 15 kg is three times less than the acceleration of a point with a mass of 5 kg.
Is it possible to solve problems of dynamics using equilibrium equations?
The amount of movement of a material point is called a vector quantity mv, equal to the product of the mass of the point and the vector of its velocity. Vector mV attached to a moving point.
Quantity of system movement is called a vector quantity Q, equal to the geometric sum (principal vector) of the momentum of all points of the system:
Vector Q is a free vector. In the SI system of 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 velocities of the points of a rolling wheel shown in Fig. 6.21), and therefore the direct summation of the vectors on the right side of equality (17.2) is difficult. Let us find a formula with the help of which the quantity Q much easier to calculate. It follows from equality (16.4) that
Taking the time derivative of both parts, we get
Hence, taking into account equality (17.2), we find that
i.e., the amount of motion of the system is equal to the product of the mass of the entire system and the speed of its center of mass.
Note that the vector Q, like the main vector of forces in statics, is some generalized vector characteristic of the motion of the entire mechanical system. In the general case of motion of a system, its momentum is 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 stationary, then the momentum of the system will be equal to zero. Such, for example, is the momentum of a body rotating around a fixed axis passing through its center of mass.
Example. Determine the amount of motion of the mechanical system (Fig. 17.1, a), consisting of cargo BUT weight t A - 2 kg, homogeneous block AT weighing 1 kg and wheels D weight mD-4 kg. Cargo BUT moving at a speed V A - 2 m/s, wheel D rolls without slipping, the thread is inextensible and weightless. Decision. The amount of movement of the body system
Body BUT moving forward and Q A \u003d m A V A(numerically Q A= 4 kg m/s, vector direction Q A coincides with the direction VA). Block AT performs rotational motion around a fixed axis passing through its center of mass; hence, QB- 0. Wheel D makes a plane-parallel
motion; its instantaneous center of velocities is at the point To, so the speed of its center of mass (points E) is equal to V E = V A /2= 1 m/s. Number of wheel movement Q D - m D V E - 4 kg m/s; vector QD directed horizontally to the left.
Depicting vectors Q A and QD in fig. 17.1, b, find the momentum Q systems according to formula (a). Considering directions and numerical values values, we get Q ~^Q A +Q E=4l/2~kg m/s, vector direction Q shown in fig. 17.1, b.
Given 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 a 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 the forces applied to the material point.
If both sides of the equation are multiplied by dt, then we get
The vector value on the right side of this equality characterizes the action exerted on the body by force in an elementary period of time dt this value is denoted dS and call 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 in modulus and in direction, then S = F(t| -/0) and S- F(t l -/ 0). In the general case, the modulus of the force impulse can be calculated from its projections onto the coordinate axes:
Now, integrating both parts of equality (17.5) with t= const, we get
Equation (17.9) expresses the theorem on changing the momentum of a point in finite (integral) form: the change in the momentum of a point over a certain period of time is equal to the momentum of the force acting on the point (or the momentum 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 onto the coordinate axes
Now consider mechanical system, consisting of P material points. Then, for each point, we can apply the momentum change theorem in the form (17.4), taking into account the external and internal forces applied to the points:
Summing up these equalities and taking into account that the sum of derivatives is equal to the derivative of the sum, we obtain
Since by property internal forces H.F.k=0 and by definition of momentum ^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, Q0 - the amount of motion of the system at times, 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 over any time is equal to the sum of the impulses of all external forces acting on the system over the same time.
In projections onto the coordinate axes, we get
From the theorem on the change in the momentum of the system, the following important consequences can be obtained, which express law of conservation of momentum of the system.
- 1. If the geometric sum of all external forces acting on the system is equal to zero (LF k=0), then from equation (17.11) it follows that in this case Q= const, i.e. the momentum vector 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 on any axis is zero (for example, I e kx = 0), then from equations (17.12) it follows 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 on the change in the momentum of the system. These forces, although they affect the momentum of individual points of the system, cannot change the momentum of the system as a whole. Given this circumstance, when solving problems, it is expedient to choose the system under consideration so that the unknown forces (all or part of them) are internal.
The law of conservation of momentum is convenient to apply in cases where the change in the speed of one part of the system is necessary to determine the speed of another part of it.
Problem 17.1. To trolley weighing t x- 12 kg moving on a smooth horizontal plane, at a point BUT a weightless rod is attached with the help of a cylindrical hinge AD length /= 0.6 m with load D weight t 2 - 6 kg at the end (Fig. 17.2). At time / 0 = 0, when the speed of the trolley and () - 0.5 m/s, rod AD starts to rotate around the axis BUT, perpendicular to the plane drawing, according to the law φ \u003d (tg / 6) (3 ^ 2 - 1) rad (/- in seconds). Define: u=f.
§ 17.3. 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 , we get
If mass M system is constant, we get
where and with - acceleration of the center of mass of the system.
Equation (17.15) 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 onto the axes of the Cartesian coordinate system.
Let's discuss the results. Let us first recall that the center of mass of the system is 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 the equations (17.15) of the theorem on the motion of the center of mass and the equation (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 as 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 by 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 moves forward, then the kinematic characteristics of all points of the body and its center of mass are the same. So a progressively moving body can always be considered as a material point with a mass equal to the mass of the entire 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 influence the movement of the center of mass in those cases when 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 law of conservation 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 it follows from equation (17.15),
what about a c = 0 or V c = const, i.e. the center of mass of this system
moves with a constant speed in magnitude and direction (otherwise, uniformly and rectilinearly). In a special case, if at the beginning the center of mass was at rest ( Vc=0), then it will remain at rest; where
track predicts that its position in space will not change, i.e. rc = const.
2. If the external forces acting on the system are such that the sum of their projections on some axis (for example, the axis X) zero (?F e kx= 0), then from equation (17.16) it follows that in this case x s=0 or V Cx \u003d x c \u003d const, i.e., the projection of the velocity of the center of mass of the system onto this axis is a constant value. In a special case, if at the initial moment Vex= 0, then at any subsequent time this value will be preserved, and hence it follows that the coordinate x s the center of mass of the system will not change, i.e. x s - const.
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; internal forces cannot change the position of the center of mass. 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 friction forces between the soles of his shoes and the road surface. With the strength of his muscles (internal forces), a person pushes off the road surface with his feet, which causes a friction force (external for a person) at the points of contact with the road, directed in the direction of his movement.
- 2. The car moves in the same way. The internal pressure forces in its engine make the wheels rotate, but since the latter have traction, the friction forces that arise “push” the car forward (as a result, the wheels do not rotate, but move in a plane-parallel way). 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, i.e., rotate.
- 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 momentum of this system. However, each of the parts of this system will move, for example, the boat forward, and the water that the oars throw back.
- 4. In airless space, when the rocket is moving, the “discarded mass” should be “taken with you”: the jet engine informs the rocket about the movement by throwing back the combustion products of the fuel that the rocket is filled with.
- 5. When descending on a parachute, you can control the movement of the center of mass of the man-parachute system. If by muscular effort a person pulls the parachute lines in such a way that the shape of its canopy or the angle of attack of the air flow changes, then this will cause a change in the external influence of the air flow, and thereby affect the movement of the entire system.
Problem 17.2. AT task 17.1 (see Figure 17.2) determine: 1) law of motion of the trolley X (= /)(/), if it is known that at the initial moment of time t 0 = About 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).
Decision. Here, as in problem 17.1, we consider a system consisting of a trolley and a load D, in an arbitrary position under the action of external forces applied to it (see Fig. 17.2). Coordinate axes Ohu draw so that the x-axis is horizontal and the x-axis at passed through the point A 0 , i.e. the location of the point BUT at the time t-t 0 - 0.
1. Determination of the law of motion of the cart. To determine x, = /, (0, we use the theorem on the motion of the center of mass of the system. Let us compose a 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 \u003d B, i.e., the projection of the velocity of the center of mass of the system on the x-axis is a constant value. Since at the initial moment of time
Integrating the equation Mx s= 0, we get
i.e. coordinate x s the center of mass of the system is constant.
Let's write the expression Mx s 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 s - t(x( + t 2 x 2".
for an arbitrary point in time
for time point / () = 0, X (= 0 and
In accordance with equality (b), the coordinate x s the center of mass of the entire system remains unchanged, i.e. x c (t). Therefore, by equating expressions (c) and (d), we obtain the dependence of the x coordinate on time.
Answer: X - 0.2 m, where t- in seconds.
2. Reaction definition 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", we get
According to the formula that determines the ordinate u s the center of mass of the system, Mu s = t (y x + t 2 y 2, where y, = at C1,at 2= yD = Ata ~ 1 cos Ф» we get
Differentiating this equality twice with respect to time (taking into account that at C1 and at A the quantities are constant and, consequently, their derivatives are equal to zero), we find
Substituting this expression into equation (e), we determine the required dependence N from t.
Answer: N- 176,4 + 1,13,
where φ \u003d (i / 6) (3 / -1), t- in seconds N- in newtons.
Problem 17.3. Electric motor mass t x attached to the horizontal surface of the foundation with bolts (Fig. 17.3). On the motor shaft at right angles to the axis of rotation, a weightless rod of length l is fixed at one end, and a point weight is mounted on the other end of the rod BUT weight t 2 . The shaft rotates uniformly at an angular velocity o. Find the horizontal pressure of the motor on the bolts. Decision. Consider a mechanical system consisting of a motor and a point weight BUT, in an arbitrary position. Let us depict the external forces acting on the system: gravity R x, R 2, foundation reaction in the form of a vertical force N and horizontal force R. 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 on the change in the momentum of the system in projection onto the horizontal axis x:
For the system under consideration in its arbitrary position, given that the amount of motion of the motor housing is zero, we obtain Qx = - t 2 U A col. Taking into account that V A = a s/, φ = ω/ (uniform rotation of the motor), we get Q x - - m 2 co/cos co/. differentiating Qx 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 forcing (see § 14.3), when they act, forced vibrations of structures occur.
Exercises for independent work
- 1. What is called the momentum of a point and a mechanical system?
- 2. How does the momentum of a point moving uniformly around 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 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 uniformly 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. In which direction will the boat move if they switch places?
In which case will the displacement module of the boat be large: 1) if the child goes to the adult in the stern; 2) if an adult goes to the 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?
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).
Since the mass of the point is constant, and its acceleration, equation (2), which expresses the basic law of dynamics, can be represented as
Equation (32) simultaneously expresses the theorem on the change in the momentum of a point in differential form: the time derivative of the momentum of a point is equal to the sum of the forces acting on the point
Let the moving point have a speed at a moment of time and a speed at a moment. Let us then multiply both parts of equality (32) by and take certain integrals from them. In this case, on the right, where the integration is over time, the limits of the integral will be and on the left, where the speed is integrated, the limits of the integral will be the corresponding values of the speed
Since the integral of is equal, as a result we get
The integrals on the right, as follows from formula (30), represent the impulses of the acting forces. Therefore, it will finally
Equation (33) expresses the theorem on the change in the momentum of a point in its final form: the change in the momentum of a point over a certain period of time is equal to the sum of the impulses of all forces acting on the point over the same period of time.
When solving problems, instead of the vector equation (33), equations in projections are often used. Projecting both parts of equality (33) onto the coordinate axes, we obtain
In the case of rectilinear motion occurring along the axis, the theorem is expressed by the first of these equations.
Problem solving. Equations (33) or (34) allow, knowing how its speed changes when a point moves, to determine the impulse of the acting forces (the first problem of dynamics) or, knowing the impulses of the acting forces, to determine how the speed of the point changes when moving (the second problem of dynamics). When solving the second problem, when the forces are given, it is necessary to calculate their momenta. As can be seen from equalities (30) or (31), this can be done only when the forces are constant or depend only on time.
Thus, equations (33), (34) can be directly used to solve the second problem of dynamics, when the number of data and sought values in the problem includes: active forces, the time of movement of the point and its initial and final velocities (i.e., the quantities ), and the forces must be constant or depend only on time.
Problem 95
Decision. According to the theorem on the change in the momentum of the System, geometrically the difference between these quantities of motion (Fig. 222), we find from the resulting right triangle
But according to the conditions of the problem, therefore,
For an analytical calculation, using the first two of equations (34), we can find
Problem 96. A load that has a mass and lies on a horizontal plane is informed initial speed The subsequent movement of the load is decelerated by a constant force F. Determine how long the load will stop,
Decision. According to the problem data, it is clear that the proven theorem can be used to determine the time of motion. We depict the load in an arbitrary position (Fig. 223). It is affected by the force of gravity P, the reaction of the plane N and the braking force F. By directing the axis in the direction of motion, we compose the first of equations (34)
In this case - the speed at the moment of stopping), and . Of the forces, only the force F gives the projection onto the axis. Since it is constant, where is the braking time. Substituting all these data into equation (a), we get the desired time from