Dynamics:
Dynamics material point
§ 28. Theorem on the change in the momentum of a material point. Theorem on the change in the angular momentum of a material point
Problems with solutions
28.1 A railway train moves along a horizontal and straight section of track. When braking, a resistance force equal to 0.1 of the train weight develops. At the start of braking, the speed of the train is 20 m/s. Find the braking time and stopping distance.
SOLUTION
28.2 On a rough inclined plane making an angle α=30° with the horizon, a heavy body descends without initial velocity. Determine during what time T the body will pass the way length l=39.2 m, if the coefficient of friction f=0.2.
SOLUTION
28.3 A train of mass 4*10^5 kg enters the rise i=tg α=0.006 (where α is the angle of rise) at a speed of 15 m/s. The coefficient of friction (coefficient of total resistance) when the train is moving is 0.005. 50 s after the train enters the rise, its speed drops to 12.5 m/s. Find the traction force of the locomotive.
SOLUTION
28.4 A weight M is tied to the end of an inextensible string MOA, part of which OA is passed through a vertical tube; the weight moves around the axis of the tube along a circle of radius MC=R, making 120 rpm. Slowly pulling the thread OA into the tube, shorten the outer part of the thread to the length OM1, at which the weight describes a circle with a radius R/2. How many revolutions per minute does the weight make along this circle?
SOLUTION
28.5 To determine the mass of a loaded train, a dynamometer was installed between the diesel locomotives and wagons. The average reading of the dynamometer for 2 minutes turned out to be 10 ^ 6 N. During the same time, the train picked up a speed of 16 m / s (at first the train stood still). Find the mass of the composition if the coefficient of friction f=0.02.
SOLUTION
28.6 What should be the coefficient of friction f of the wheels of a braked car on the road if, at a driving speed v = 20 m/s, it stops 6 s after the start of braking.
SOLUTION
28.7 A bullet of mass 20 g flies out of the barrel of a rifle with a speed of v=650 m/s, running through the bore in the time t=0.00095 s. Determine the average pressure of the gases ejecting the bullet if the cross-sectional area of the channel is σ=150 mm^2.
SOLUTION
28.8 The point M moves around a fixed center under the influence of the force of attraction to this center. Find the velocity v2 at the point of the trajectory furthest from the center, if the velocity of the point at the position closest to it is v1=30 cm/s, and r2 is five times greater than r1.
SOLUTION
28.9 Find the momentum of the resultant of all forces acting on the projectile during the time when the projectile moves from the initial position O to the highest position M. Given: v0=500 m/s; α0=60°; v1=200 m/s; projectile weight 100 kg.
SOLUTION
28.10 Two asteroids M1 and M2 describe the same ellipse, in the focus of which S is the Sun. The distance between them is so small that the arc M1M2 of the ellipse can be considered a straight line segment. It is known that the length of the M1M2 arc was a when its middle was at perihelion P. Assuming that the asteroids move with equal sectorial velocities, determine the length of the M1M2 arc when its middle passes through aphelion A, if it is known that SP=R1 and SA =R2.
SOLUTION
28.11 A boy of mass 40 kg stands on the runners of a sports sleigh, the mass of which is 20 kg, and every second he pushes with an impulse of 20 N * s. Find the speed acquired by the sleigh in 15 s if the coefficient of friction f=0.01.
SOLUTION
28.12 The point makes a uniform movement along a circle with a speed v=0.2 m/s, making a complete revolution in time T=4 s. Find the momentum S of the forces acting on the point during one half-cycle if the mass of the point is m=5 kg. Determine the average value of the force F.
SOLUTION
28.13 Two mathematical pendulums suspended on threads of lengths l1 and l2 (l1>l2) oscillate with the same amplitude. Both pendulums simultaneously began to move in the same direction from their extreme deflected positions. Find the condition that the lengths l1 and l2 must satisfy in order for the pendulums to simultaneously return to the equilibrium position after a certain period of time. Determine the smallest time interval T.
SOLUTION
28.14 A ball of mass m, tied to an inextensible thread, slides along a smooth horizontal plane; the other end of the thread is pulled at a constant speed a into a hole made on a plane. Determine the movement of the ball and the tension of the thread T, if it is known that at the initial moment the thread is located in a straight line, the distance between the ball and the hole is R, and the projection of the initial velocity of the ball onto the perpendicular to the direction of the thread is v0.
SOLUTION
28.15 Determine the mass M of the Sun, having the following data: the radius of the Earth R=6.37*106 m, the average density is 5.5 t/m3, the semi-major axis of the earth's orbit a=1.49*10^11 m, the time of the Earth's revolution around the Sun T=365.25 days Strength gravity between two masses equal to 1 kg, at a distance of 1 m, we consider equal to gR2/m H, where m is the mass of the Earth; It follows from Kepler's laws that the force of attraction of the Earth by the Sun is equal to 4π2a3m/(T2r2), where r is the distance of the Earth from the Sun.
SOLUTION
28.16 A point of mass m subjected to the action of a central force F describes a lemniscate r2=a cos 2φ, where a is a constant value, r is the distance of the point from the center of force; at the initial moment r=r0, the speed of the point is equal to v0 and makes an angle α with the straight line connecting the point with the center of force. Determine the magnitude of the force F, knowing that it depends only on the distance r. According to Binet's formula, F =-(mc2/r2)(d2(1/r)/dφ2+1/r), where c is twice the sector velocity of the point.
SOLUTION
28.17 A point M, whose mass is m, moves near a fixed center O under the influence of a force F emanating from this center and depending only on the distance MO=r. Knowing that the speed of the point is v=a/r, where a is a constant value, find the magnitude of the force F and the trajectory of the point.
SOLUTION
28.18 Determine the movement of a point whose mass is 1 kg, under the action of a central force of attraction, inversely proportional to the cube of the distance of the point from the center of attraction, with the following data: at a distance of 1 m, the force is 1 N. At the initial moment, the distance of the point from the center of attraction is 2 m, velocity v0=0.5 m/s and makes an angle of 45° with the direction of the straight line drawn from the center to the point.
SOLUTION
28.19 A particle M of mass 1 kg is attracted to a fixed center O by a force inversely proportional to the fifth power of the distance. This force is equal to 8 N at a distance of 1 m. At the initial moment, the particle is at a distance of OM0=2 m and has a velocity perpendicular to OM0 and equal to 0.5 m/s. Determine the trajectory of the particle.
SOLUTION
28.20 A point of mass 0.2 kg, moving under the influence of the force of attraction to a fixed center according to Newton's law of gravitation, describes a complete ellipse with semi-axes 0.1 m and 0.08 m for 50 s. Determine the largest and smallest values of the attractive force F during this movement.
SOLUTION
28.21 A mathematical pendulum, each swing of which lasts one second, is called a second pendulum and is used to measure time. Find the length l of this pendulum, assuming the acceleration due to gravity to be 981 cm/s2. What time will this pendulum on the Moon show, where the acceleration of gravity is 6 times less than the earth? What length l1 should a lunar second pendulum have?
SOLUTION
28.22 At some point on the Earth, the second pendulum counts time correctly. Being moved to another location, it lags behind by T seconds per day. Determine the acceleration due to gravity in the new position of the second pendulum.
Of the two main dynamic characteristics, the quantity is vector. Sometimes, when studying the motion of a point, instead of changing the vector itself, it turns out to be necessary to consider a change in its momentum. Moment of a vector about a given center ABOUT or axes z denoted by or and are named respectively angular momentum or angular momentum points about this center (axis). The moment of the vector is calculated in the same way as the moment of force. In this case, the vector is considered to be attached to the moving point. Modulo , where h- length of a perpendicular dropped from the center ABOUT to the direction of the vector (Fig. 15).
Theorem of moments about the center. Let us find for a material point moving under the action of a force F(Fig.15), the relationship between the moments of the vectors and relative to some fixed center ABOUT. At the end it was shown that .
Similarly 
In this case, the vector is directed perpendicular to the plane passing through the center ABOUT and vector , and vector - perpendicular to the plane passing through the center ABOUT and vector .
Fig.15
Differentiating the expression with respect to time, we get:
But , as the cross product of two parallel vectors, a . Hence,
As a result, we have proved the following theorem of moments about the center: the time derivative of the angular momentum of a point, taken relative to some fixed center, is equal to the moment of the force acting on the point relative to the same center
.
A similar theorem holds for the moments of the vector
forces about some axis z, which can be seen by projecting both sides of the equality 
to this axis. The mathematical expression of the theorem of moments about an axis is given by the formula 
.
Questions for self-examination
What are the two measures of mechanical motion and their corresponding force meters?
What are the driving forces?
What forces are called resistance forces?
Write down the formulas for determining work in translational and rotational movements?
What is called circumferential force? What is torque?
Formulate a theorem on the work of the resultant.
How is the work of a constant modulus and direction of force on rectilinear movement determined?
What is the work of the sliding friction force if this force is constant in magnitude and direction?
How in a simple way it is possible to calculate the work of a constant modulo and direction of force on a curvilinear movement?
What is the work of the resultant force.
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?
What is the vector expression for elementary work?
What is the expression of the elementary work of the force through the projection of the force on the coordinate axes?
Write different types of a curvilinear integral that determines the work of a variable force on a finite curvilinear displacement.
What is the graphical method for determining the work of a variable force on a curvilinear movement?
How are the work of gravity and the work of elastic force calculated?
On what displacements is the work of gravity: a) positive, b) negative, c) equal to zero.
In what case is the work of the elastic force positive and in what case is it negative?
What force is called: a) conservative; b) non-conservative; c) dissipative?
What is called the potential of conservative forces?
What field is called potential?
What is a power function?
What is a force field? Give examples of force fields.
What mathematical dependencies are the field potential and the force function related to?
How to determine the elementary work of the forces of the potential field and the work of these forces on the final displacement of the system, if the force function of the field is known?
What is the work of the forces acting on the points of the system in a potential field, on a closed displacement?
What is the potential energy of the system in any of its positions?
What is the change in the potential energy of a mechanical system when it is moved from one position to another?
What relationship exists between the force function of a potential field and the potential energy of a system located in this field?
Calculate the change in kinetic energy of a point with a mass of 20 kg if its speed increased from 10 to 20 m/s?
How are the projections on the coordinate axes of the force acting in the potential field on any point of the system determined?
What surfaces are called equipotential and what are their equations?
How is the force acting on a material point in a potential field directed relative to the equipotential surface passing through this point?
What is the potential energy of a material point and a mechanical system under the action of gravity?
What form do the equipotential surfaces of the gravity field and the Newtonian gravitational force have?
What is the law of conservation and transformation of mechanical energy?
Why does a material point describe a plane curve under the action of a central force?
What is called sector speed and how to express its module in polar coordinates?
What is the area law?
What kind of differential equation in the Binet form, which determines the trajectory of a point moving under the action of a central force?
What is the formula for the Newtonian gravitational force?
What is the canonical form of the conic section equation and at what values of eccentricity does the trajectory of a body moving in the field of Newtonian gravitational force represent a circle, ellipse, parabola, hyperbola?
Formulate the laws of planetary motion discovered by Kepler.
Under what initial conditions does a body become a satellite of the Earth and under what conditions is it able to overcome the earth's gravity?
What are the first and second space speeds?
Write down the formulas for calculating work during translational and rotational movements?
A wagon weighing 1000 kg is moved along a horizontal track by 5 m, the friction coefficient is 0.15. Determine the work done by gravity?
Write down the formulas for calculating the power during translational and rotational movements?
Determine the power required to lift a load of 0.5 kN to a height of 10 m in 1 min?
What is the work of the force applied to a rectilinearly moving body with a mass of 100 kg, if the speed of the body increased from 5 to 25 m/s?
Determine the overall efficiency of the mechanism if, with an engine power of 12.5 kW and a total movement resistance force of 2 kN, the movement speed is 5 m / s.
If the car enters a mountain with the same engine power, then it reduces the speed. Why?
The work of a constant force in rectilinear movement W\u003d 10 J. What angle does the direction of force make with the direction of movement?
1) acute angle;
2) right angle;
3) obtuse angle.
How will the kinetic energy of a point moving in a straight line change if its speed doubles?
1) will double;
2) will quadruple.
What is the work done by gravity during the horizontal displacement of the body?
1) the product of gravity and displacement;
2) the work of gravity is zero.
Tasks for independent solution
Task 1. A stone is thrown horizontally from a tower 25 m high with a speed of 15 m/s. Find the kinetic and potential energy of the stone one second after the start of motion. The mass of the stone is 0.2 kg.
Task 2. A stone is thrown at an angle of 60° to the horizontal with a speed of 15 m/s. Find the kinetic, potential and total energy of the stone: 1) one second after the start of movement, 2) at the highest point of the trajectory. The mass of the stone is 0.2 kg. Ignore air resistance.
Task 3.
Task 4. A tank with a mass of 15 tons and a power of 368 kW rises uphill with a slope of 30°. What is the maximum speed the tank can develop?
Task 5. A chandelier weighing 100 kg is suspended from the ceiling on a metal chain, the length of which is 5 m. What is the height to which the chandelier can be deflected so that the chain does not break during subsequent swings, if it is known that the break occurs at a tension force of 2 kN?
Task 6. The wind blowing at a speed v 0 \u003d 20 m / s acts on a sail with an area s \u003d 25 m 2 with a force F \u003d a sρ(v 0 -v) 2 /2, where a- dimensionless coefficient, ρ - air density, v - vessel speed. Determine the conditions under which the wind power is maximum. Find the work of the wind force.
Task 7. A car weighing 1 ton moves downhill with the engine off at a constant speed of 54 km/h. The slope of the mountain is 4 m for every 100 m of the path. How much power must be developed by the engine of this car in order for the car to move at the same speed uphill with the same slope?
Task 8. A hammer with a mass of 1.5 tons strikes a red-hot ingot lying on an anvil and deforms the ingot. The mass of the anvil together with the blank is 20 tons. Determine the efficiency at the hammer blow, considering the blow to be inelastic. Consider the work done during the deformation of the blank as useful.
Task 9. The striker (shock part) of a pile hammer with a mass of 500 kg falls on a pile with a mass of 100 kg at a speed of 4 m/s. Determine: a) the kinetic energy of the striker at the moment of impact; b) the energy spent on deepening the pile into the ground, c) the energy spent on the deformation of the pile, d) the efficiency of the impact of the striker on the pile. The impact of the striker on the pile is considered as inelastic.
Task 10. The projectile flies out of the gun at an angle α to the horizon with a speed v 0 . In the upper part of the trajectory, the projectile breaks into two equal parts, and the velocities of the parts immediately after the explosion are horizontal and lie in the plane of the trajectory. One half fell at a distance s from the gun in the direction of the shot. Determine the place where the second half fell, if it is known that it fell further than the first. Assume that the flight of the projectile takes place in an airless space.
Task 11. The projectile flies in airless space along a parabola and breaks at the top of the trajectory into two equal parts. One half of the projectile fell vertically downwards, the other at a distance s horizontally from the point of rupture. Determine the speed of the projectile before burst, if it is known that the explosion occurred at a height H and the half of the projectile that fell vertically downwards fell for a time τ.
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Theorem on the change in the angular momentum of a material point
Moment of momentum
The moment of momentum of the point M relative to the center O is a vector directed perpendicular to the plane passing through the momentum vector and the center O in the direction from which the rotation of the momentum vector about the center O is visible counterclockwise.
The moment of momentum of the point M relative to os and is equal to the product of the projection of the momentum vector on a plane perpendicular to the axis on the shoulder of this projection relative to the point O of the intersection of the axis with the plane.
Theorem on the change in the angular momentum of a material point relative to the center
The time derivative of the angular momentum of a material point relative to some fixed center is equal to the geometric sum of the moments of forces acting on the point relative to the same center.
Theorem on the change in the angular momentum of a material point about the axis
The time derivative of the angular momentum of a material point with respect to some fixed axis is equal to the algebraic sum of the moments of forces acting on the point with respect to the same axis.
Laws of conservation of angular momentum of a material point
- If the line of action of the resultant forces applied to the material point all the time passes through some fixed center, then the angular momentum of the material point remains constant.
- If the resultant moment of the forces applied to a material point relative to a certain axis is always equal to zero, then the angular momentum of the material point relative to the same axis remains constant.
Theorem on the change in the main moment of the momentum of the system
momentum
Kinetic moment or main moment of momentum of a mechanical system relative to the center call a vector equal to the geometric sum of the moments of the momentum of all material points of the system relative to the same center.
The angular momentum or the main moment of the momentum of a mechanical system about the axis call the algebraic sum of the moments of the momentum of all material points about the same axis
Projection angular momentum of a mechanical system relative to the center O on an axis passing through this center is equal to the angular momentum of the system relative to this axis.
Theorem on the change in the main moment of the momentum of the system (relative to the center) - the theorem of moments
The time derivative of the angular momentum of a mechanical system with respect to some fixed center is geometrically equal to the main moment of external forces acting on this system with respect to the same center
Theorem on the change in the angular momentum of a mechanical system (with respect to the axis)
The time derivative of the kinetic moment of a mechanical system with respect to some axis is equal to the main moment of external forces with respect to the same axis.
Laws of conservation of the kinetic moment of a mechanical system
- If the main moment of external forces relative to some immovable center is always equal to zero, then the kinetic moment of the mechanical system relative to this center is a constant value.
- If the main moment of external forces about a certain axis is equal to zero, then the kinetic moment of the mechanical system about the same axis is a constant.
- The moment theorem is of great importance in the study rotary motion bodies and allows not to take into account obviously unknown internal forces.
- Internal forces cannot change the main moment of momentum of the system.
Kinetic moment of a rotating system
For a system that rotates around a fixed axis (or an axis passing through the center of mass), the kinetic moment about the axis of rotation is equal to the product of the moment of inertia about this axis and the angular velocity.
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The amount of motion of the system, as a vector quantity, is determined by formulas (4.12) and (4.13).
Theorem. The time derivative of the amount of motion of the system is equal to the geometric sum of all external forces acting on it.
In the projections of the Cartesian axes, we obtain scalar equations.
You can write a vector
(4.28)
and scalar equations
Which express the theorem 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 for the same period of time. When solving problems, equations (4.27) are more often used
Law of conservation of momentum
Theorem on the change of the kinetic moment
The theorem on the change in the moment of momentum of a point relative to the center: the time derivative of the moment of momentum of a point relative to a fixed center is equal to vector moment acting on a point of force about the same center.
Or 
(4.30)
Comparing (4.23) and (4.30), we see that the moments of the vectors and are connected by the same dependence as the vectors themselves and are connected (Fig. 4.1). If we project equality onto the axis passing through the center O, we get
(4.31)
This equality expresses the theorem of the moment of momentum of a point about an axis.
|
(4.32)
If we project the expression (4.32) onto the axis passing through the center O, then we obtain an equality that characterizes the theorem on the change in the angular momentum relative to the axis.
(4.33)
By substituting (4.10) into equality (4.33), one can write the differential equation of a rotating rigid body (wheels, axles, shafts, rotors, etc.) in three forms.
(4.34)
(4.35)
(4.36)
Thus, it is advisable to use the theorem on the change in the kinetic moment to study the motion of a rigid body, which is very common in technology, its rotation around a fixed axis.
The law of conservation of the angular momentum of the system
1. Let in expression (4.32) .
Then it follows from equation (4.32) that , i.e. if the sum of the moments of all external forces applied to the system relative to a given center is zero, then the kinetic moment of the system relative to this center will be numerically and in direction will be constant.
2. If , then . Thus, if the sum of the moments of external forces acting on the system with respect to some axis is equal to zero, then the kinetic moment of the system with respect to this axis will be a constant value.
These results express the law of conservation of angular momentum.
In the case of a rotating rigid body, it follows from equality (4.34) that if , then . From here we come to the following conclusions:
If the system is immutable (absolutely solid), then , therefore, and and the rigid body rotates around a fixed axis with a constant angular velocity.
If the system is changeable, then . With an increase (then the individual elements of the system move away from the axis of rotation), the angular velocity decreases, because , and increases with decreasing, thus, in the case of a variable system, with the help of internal forces, it is possible to change the angular velocity.
Second task D2 control work is devoted to the theorem on the change in the kinetic moment of the system about the axis.
Task D2
A homogeneous horizontal platform (round with radius R or rectangular with sides R and 2R, where R = 1.2 m) with mass kg rotates with an angular velocity around the vertical axis z, which is spaced from the center of mass C of the platform at a distance OC = b (Fig. D2.0 – D2.9, table D2); dimensions for all rectangular platforms are shown in fig. D2.0a (top view).
At the moment of time, a load D with a mass of kg begins to move along the platform chute (under the action of internal forces) according to the law , where s is expressed in meters, t is in seconds. At the same time, a pair of forces with a moment M (given in newtonometers) begin to act on the platforms; at M< 0 его направление противоположно показанному на рисунках).
Determine, neglecting the mass of the shaft, the dependence i.e. platform angular velocity as a function of time.
In all figures, the load D is shown in a position where s > 0 (when s< 0, груз находится по другую сторону от точки А). Изображая чертеж решаемой задачи, провести ось z на заданном расстоянии OC = b от центра C.
Directions. Task D2 - on the application of the theorem on the change in the angular momentum of the system. When applying the theorem to a system consisting of a platform and a load, the angular momentum of the system about the z-axis is defined as the sum of the moments of the platform and the load. In this case, it should be taken into account that the absolute speed of the load is the sum of the relative and portable speeds, i.e. . Therefore, the amount of movement of this cargo 
. Then we can use the Varignon theorem (statics), according to which ; these moments are calculated in the same way as the moments of forces. The course of the solution is explained in more detail in example D2.
When solving the problem, it is useful to depict on the auxiliary drawing a view of the platform from above (from the end z), as is done in Fig. D2.0,a - D2.9,a.
The moment of inertia of a plate with mass m about the axis Cz, perpendicular to the plate and passing through its center of mass, is equal to: for a rectangular plate with sides and
;
For a round insert of radius R
| Condition number | b | s = F(t) | M |
| R R/2 R R/2 R R/2 R R/2 R R/2 | -0.4 0.6 0.8 10t 0.4 -0.5t -0.6t 0.8t 0.4 0.5 | 4t -6 -8t -9 6 -10 12 |
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Example D2. A homogeneous horizontal platform (rectangular with sides 2l and l), having a mass, is rigidly fastened to a vertical shaft and rotates with it around an axis z with angular velocity (Fig. E2a ). At the moment of time, a torque M begins to act on the shaft, directed oppositely ; at the same time cargo D mass located in the gutter AB at the point WITH, starts to move along the chute (under the action of internal forces) according to the law s = CD = F(t).
Given: m 1 \u003d 16 kg, t 2= 10 kg, l\u003d 0.5 m, \u003d 2, s \u003d 0.4t 2 (s - in meters, t - in seconds), M= kt, where k=6 Nm/s. Determine: - the law of change of the angular velocity of the platform.
Solution. Consider mechanical system, consisting of a platform and cargo D. To determine w, we apply the theorem on the change in the angular momentum of the system about the axis z:
(1)
Let us depict the acting on the system external forces: reaction gravity and torque M. Since the forces and are parallel to the z-axis, and the reactions intersect this axis, their moments about the z-axis are equal to zero. Then, considering the positive direction for the moment (i.e., counterclockwise), we obtain 
and equation (1) will take this form.
In some problems, as a dynamic characteristic of a moving point, instead of the momentum itself, its moment relative to some center or axis is considered. These moments are defined in the same way as the moments of force.
Moment of momentum material point with respect to some center O is called a vector defined by the equality
The angular momentum of a point is also called angular momentum .
Moment of momentum relative to any axis, passing through the center O, is equal to the projection of the momentum vector on this axis.
If the momentum is given by its projections 
on the coordinate axis and the coordinates of a point in space are given, then the moment of momentum relative to the origin is calculated as follows:
The projections of the angular momentum on the coordinate axes are:
The SI unit of momentum is -.
The theorem on the change in the angular momentum of a point.
Theorem. The time derivative of the moment of momentum of a point, taken with respect to some center, is equal to the moment of the force acting on the point with respect to the same center.
Proof: Differentiate moment of momentum with respect to time
, 
, Consequently , (*)
Q.E.D.
Theorem. The time derivative of the momentum of the point's momentum, taken with respect to any axis, is equal to the moment of the force acting on the point with respect to the same axis.
To prove it, it suffices to project the vector equation (*) onto this axis. For the axis it would look like this:
Consequences from the theorems:
1. If the moment of force relative to a point is equal to zero, then the moment of momentum relative to this point is a constant value.
2. If the moment of force about an axis is zero, then the moment of momentum about this axis is a constant value.
Force work. Power.
One of the main characteristics of a force that evaluates the effect of a force on a body during some movement.
Elementary work of force a scalar value equal to the product of an elementary displacement and the projection of the force on this displacement.
The SI unit of work is −
When at 
Special cases:
The elementary displacement is equal to the differential of the radius vector of the force application point.
Elementary work of force is equal to the scalar product of the force and the elementary displacement or the differential of the radius vector of the force application point.
Elementary work of force is equal to the scalar product of the elementary impulse of the force and the speed of the point.
If the force is given by its projections () on the coordinate axes and the elementary displacement is given by its projections () on the coordinate axes, then the elementary work of the force is equal to:
(analytical expression for elementary work).
The work of a force on any finite displacement is equal to the integral of the elementary work taken along this displacement.
By the power of strength is the quantity that determines the work done by the force per unit time. In general, power is equal to the first time derivative of work.
, 
Power is equal to the scalar product of force and speed.
The SI unit of power is − 
In technology, the unit of force is taken.
Example 1. The work of gravity.
Let the point M, which is affected by the force of gravity P, move from the position 
into position. We choose the coordinate axes so that the axis is directed vertically upwards.
Then, , , and
The work of gravity is equal to the product of the modulus of force and the vertical displacement of the point of its application, taken with a plus or minus sign. The work is positive if the start point is higher than the end point, and negative if the start point is lower than the end point.
Example 2. The work of the elastic force.
Consider a material point fixed on an elastic stiffener c, which oscillates along the x axis. Elastic force (or restoring force). Let the point M, which is affected only by the elastic force, move from position to position. ( , ).
The power of a pair of forces is equal to
Kinetic energy of a point
Kinetic energy of a material point (or its living force) is called half the product of the mass of a point and the square of its speed.
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