Movement along the horizontal surface of the formula. Movement on a horizontal plane

So far we have considered the movement of a top with one fixed point, the presence of which essentially caused precessional and nutational movements. How will the top behave if there is no such point and it can move freely on a horizontal surface? This problem is considered in the books, where a semi-qualitative explanation of the nature of the top’s movement is given. We will give our explanation, although also approximate.
Let us analyze the case considered in the work, when the top is on an absolutely smooth surface, that is, there is no friction between the surface and the top. If a rotating top is carefully placed on the surface at an angle to the vertical without pushing, then its end in contact with the surface will describe figures characteristic of a combination of nutational and precessional movements (Fig. 1). This nature of the movement of the top can be explained by the following reasons.
1. Affects the top active moment forces G and N, equal in magnitude to each other. Under the influence of this moment, as in previous examples, the top will begin to perform precessional and nutation movements with support on the tip. The law of this motion can be approximately calculated if the top of the top is considered motionless.
2. Since there is no friction between the tip of the top and the surface, the movement of the center of mass of the top will lead to the movement of its top relative to the surface, and minor movements of the center of mass vertically will lead to a significant change in the angle a(see Fig. 1,b). Let's make basic calculations to determine the ratio Dx/Dz. First let's find the angles a1 And a2. From Figure 1, b it follows:
; (1)
, (2)
Where ls- the distance from the point of contact to the center of mass of the top, from where we get:
(3)
(4)
Now let's find interdependent changes in coordinates Dx And Dz:
(5)
(6)
Then the ratio of increments DX/ DZ will be determined by the expression:
(7)
At angle a1= 100 attitude Dx/ Dz varies from 5 to 3.5 when changing Dz/z1 from 0.01 to 0.05. In addition, the radius value OK1 is approximately 0.18 of the coordinate length Z1. As a result, minor fluctuations of the center of mass relative to it initial position will seem to intensify and will be clearly visible on the surface. The work states that the center of mass will be stationary, but this cannot be, since the end of the top must then break away from the surface.
3. The nutational vibrations of the top create stability in its movement and prevent it from falling to the surface.
The picture of the top's motion will be even more complex if it moves along a surface in the presence of friction. If a rotating top is given horizontal speed by a push, it will begin to move in a converging spiral (see Fig. 2). This will allow light tops to move along the polished surface. After several revolutions along this spiral, the top will stop at the point ABOUT and will continue to rotate around its axis, staying in one place.
So what is the reason that makes the top move in a spiral rather than in a straight line?
Let's look at this question in general outline, since the physical picture will be quite complex. The main reason for this behavior of the top is the force of friction Ftr between the top and the surface. The friction force will slow down the movement, resulting in a force of inertia applied at the center of mass of the top and directed in the direction of movement. Under the influence of inertia force, creating an overturning moment My, the axis of rotation of the top will tilt forward at a certain angle a and take a position Z', and the center of mass S- position S'(see Fig. 3, a, b). When the rotating top is turned, it comes into action gyroscopic effect, considered by us in §5, as a result of which the moment arises Mx, rotating the top around an axis X. To determine the direction of the moment Mx Let us consider the velocity picture that appears when adding the velocities of a rotating top Vr at any point and equal to the product w per radius r and speed DVr from rotation of the top around its axis Y(see Fig. 3,c). As a result of the addition of velocities in an arbitrary section of the top, the instantaneous center of velocities Pv from the axis of the top is shifted to another point. As a result, there will be Reactive force inertia F, which will cause the top to move to the new position of the point Pv, so the top will begin to rotate around its axis X counterclockwise when viewed from the end of this axis. The magnitude of the torque in accordance with formula (5.16) is determined by the expression:
, (8)
Where Jx is the moment of inertia of the top relative to the X axis passing through the center of mass of the top.
As a result of rotation around the X axis, the center of mass of the top will take the position S'', and the axis Z'- position Z'', turning to the corner b(see Fig. 3, a, b). The resulting movement of the center of mass of the top will be determined by the segment DZ, equal geometric sum movements DX And DY. Thus, the center of mass of the top will shift relative to the coordinate system X,Y,Z, the beginning of which is at the point A, and will lie on a straight line I-I, located at an angle g to the axis X.
Under the influence of moments My And Mx the top should have fallen, but here the gyroscopic effect due to the weight of the top again manifests itself G. We examined this effect in detail in §§ 4-7, so we will simply indicate the direction of the arising periodic forces of inertia and. To do this, we show the section I-I top with a vertical plane passing through the Z axis (see Fig. 3,d), and then a section II-II plane perpendicular to the axis Z'' and passing through the center of mass of the top (see Fig. 3,e). The magnitude of these forces is determined by the expressions:
; (9)
, (10)
Where y- angle between axes Z'' and Z.
These forces will influence the movement of the top, causing it to make additional movements along the surface. These movements will be determined by the projections of forces in the horizontal direction (see Fig. 3d):
; (11)
(12)
It should be noted that after one revolution of the top around its axis, the resulting displacement from the action of the force will be equal to zero, and the resulting displacement along the axis Y depending on the force will be determined by its projection onto the axis Y and it will be equal:
(13)
those. the top will move by this amount along the surface in the direction of the axis Y for one revolution under the influence of inertial forces.
As a result of the action of all factors: the initial push and the inertial forces that appear, the top will move along curvilinear trajectory, which we will approximately consider to be an arc of a circle. Figure 4 shows the movement of the top from the initial zero to the first position after the first revolution around its axis. The amount of displacement is determined by formula (13), arc length S0S1 can be found by solving differential equation top movements:
, (14)
where V is the linear speed of movement of the top along the trajectory.
Bearing in mind that the initial speed of movement of the top along the trajectory V0, and the displacement S along the X axis is zero, we obtain the following expressions:
; (15)
, (16)
where m is the mass of the top.
Based on Coulomb’s law, we represent the friction force in the form:
, (17)
Where G- weight of the top, f- sliding friction coefficient for the top-support pair of materials.
Then expressions (15) and (16) are transformed to the form:
; (18)
(19)
Since the time of one revolution of the top is equal to:
, (20)
then the speed and displacement in the first position will be respectively equal:
; (21)
(22)
Let's find the radius of curvature of the top's trajectory by replacing the arc S0S1 chord. Then we get:
(23)
Since from Figure 4 it follows that:
, (24)
expression (23) will take the form:
(25)
After determining the first position of the top, you can proceed to determining its second position by taking the first position as the initial one and entering new system coordinates In this way, step by step, you can find the entire trajectory of the top’s movement.
For step-by-step calculation of the trajectory, more convenient formulas can be derived. Let us take two adjacent positions of the top on the trajectory, separated by the time of its one revolution around the axis: positions i and i+1 (see Fig. 5). The value of velocities and displacements at these points can be found using expressions (18) and (19):
; (26)
; (27)
; (28)
(29)
The movement of the top along its trajectory between these two positions is determined by the difference in displacement Si+1 And Si:
(30)
Here: Dti- time of one revolution of the top in the i-th position, equal to:
, (31)
Where wi- angular speed of rotation of the top in the i-th position.
The angular velocity of rotation of the top as it moves along the trajectory continuously decreases due to friction on the surface and energy losses due to inertial movement due to the action of forces Fx And Fy.
To determine the angular velocity of the top in any of its positions, we write the energy balance equation:
, (32)
Where J- moment of inertia of the top relative to its axis of rotation, DAi- total energy losses during movement up to i-th position.
From expression (32) it follows:
(33)
Then the radius of curvature of the trajectory is determined by the expression:
(34)
and the angle mi using the formula:
(35)
Since the trajectory of the top’s movement is curvilinear, another force will act on the top, which will also affect the nature of the top’s movement - this centrifugal force inertia (Fig. 6):
, (36)
Where wi- angular velocity of rotation of the center of mass of the top around its axis Oi(instantaneous velocity center):
(37)
Under the influence of all forces, the top will move along a trajectory with an axis inclined relative to the vertical own rotation. And this leads to the fact that in the presence of friction, the top will roll over the surface like a conical body in the direction opposite to rotation around the point ABOUT With angular velocity w. Together with this movement the point will move A, lying on the axis of the top, as a result of which the trajectory will deviate from the circle of radius r(see Fig. 7). This is explained by the fact that the tip of the top is blunted and can be considered as part spherical surface radius rsf. As a result of rolling, the top will move away from the center of curvature of the trajectory, and its radius r will increase accordingly. This circumstance will also have a significant impact on the nature of the top’s movement. In Figure 7 rext- this is an increase in the radius of curvature of the trajectory due to the tilt of the top axis. Experiments show that at a certain initial inclination of the top's axis from the vertical, after a push the top can move in a straight line and even in a spiral twisted in the other direction.
Let's calculate the amount of movement Sk due to the top rolling relative to the point O1 for one revolution around its axis (see Fig. 8).
Linear speed moving the touch point Ak when the top rolls due to its rotation around its axis along the surface, as well as the speed and points A (the speeds of these points will be the same, since they are at the same distance from vertical axis Z1, around which rolling occurs) will be equal to:
, (38)
Where rk- radius of the conical part, which can be found from the radius of the sphere (see Fig. 8):
(39)
Magnitude linear movement points Ak will be determined by its speed:
, (40)
Where t- movement time. For one revolution of the top ( tob=2p/w), move Sk, will be equal to:
(41)
Because of this rolling, the trajectory of the top will change somewhat and the point Ak instead of the position it will fall into the position (see Fig. 9), which will change the radius of curvature of the trajectory. In accordance with Figure 9 we have:
(42)
where:
; (43)
Corner j'' can be expressed through an angle j', equating the chords and with some assumption:
; (44)
Where:
, (45)
from where we get:
(46)
Here S- movement of the top along the trajectory during one revolution.
Thus, we have examined in general terms the nature of the movement of the top when it moves along a horizontal surface, taking into account the influence of friction forces. It is interesting to note the following experimental fact: after the cessation of movement along the trajectory, the axis of rotation of the top assumes a vertical position. This phenomenon can be explained by the fact that the inertial force caused by resistance from friction forces disappears.
From the problem considered we can make the following conclusions:
1. The movement of the top after the push occurs without the influence of active external forces except for its weight. The friction force is a passive force that slows down movement.
2. The observed movement of the top along the trajectory can only be explained joint action frictional forces and inertia forces after imparting linear horizontal speed to the top V0. This is another example confirming the reality of inertial forces.

Single State exam in physics, 2009,
demo version

Part A

A1. The figure shows a graph of the projection of body velocity versus time. The graph of the projection of body acceleration versus time in the time interval from 12 to 16 s coincides with the graph

1)
2)
3)
4)

Solution. The graph shows that in the time interval from 12 to 16 s, the speed changed uniformly from –10 m/s to 0 m/s. The acceleration was constant and equal

The acceleration graph is shown in the fourth figure.

Correct answer: 4.

A2. Strip magnet with mass m brought to a massive steel plate weighing M. Compare the force of the magnet on the plate with the force of the plate on the magnet.

Solution. According to Newton's third law, the force with which the magnet acts on the plate is equal to the force with which the plate acts on the magnet.

Correct answer: 1.

A3. When moving on a horizontal surface, a sliding friction force of 10 N acts on a body weighing 40 kg. What will the sliding friction force be after reducing the body's mass by 5 times, if the friction coefficient does not change?

Solution. If your body weight decreases by 5 times, your body weight will also decrease by 5 times. This means that the sliding friction force will decrease by 5 times and amount to 2 N.

Correct answer: 2.

A4. A car and a truck are moving at speeds And . Car weight m= 1000 kg. What is the mass of the truck if the ratio of the truck's momentum to the car's momentum is 1.5?

Solution. The momentum of the car is . The truck's momentum is 1.5 times greater. The mass of the truck is .

Correct answer: 1.

A5. Mass sled m pulled uphill with constant speed. When the sled rises to the top h from the initial position, their total mechanical energy

Solution. Since the sled is pulled at a constant speed, it kinetic energy does not change. Change complete mechanical energy sled is equal to the change in their potential energy. Total mechanical energy will increase by mgh.

Correct answer: 2.

Solution. The wavelength ratio is inversely proportional to the frequency ratio: .

Correct answer: 4.

A7. The photograph shows a setup for studying uniformly accelerated sliding of a carriage (1) weighing 0.1 kg along an inclined plane set at an angle of 30° to the horizontal.

At the moment the movement begins, the upper sensor (A) turns on the stopwatch (2), and when the carriage passes the lower sensor (B), the stopwatch turns off. The numbers on the ruler indicate the length in centimeters. What expression describes the dependence of carriage speed on time? (All values ​​are in SI units.)

Solution. From the figure it can be seen that during the time t= 0.4 s the carriage has traveled the distance s= 0.1 m. Since the initial speed of the carriage is zero, its acceleration can be determined:

.

Thus, the speed of the carriage depends on time according to the law.

Correct answer: 1.

A8. When decreasing absolute temperature of a monatomic ideal gas by 1.5 times the average kinetic energy of the thermal motion of its molecules

Solution. The average kinetic energy of thermal motion of molecules of an ideal gas is directly proportional to the absolute temperature. When the absolute temperature decreases by 1.5 times, the average kinetic energy will also decrease by 1.5 times.

Correct answer: 2.

A9. The hot liquid cooled slowly in the glass. The table shows the results of measuring its temperature over time.

There was a substance in the glass 7 minutes after the start of measurements

Solution. The table shows that in the period between the sixth and tenth minutes the temperature in the glass remained constant. This means that at this time crystallization (solidification) of the liquid took place; the substance in the glass was simultaneously in both liquid and solid states.

Correct answer: 3.

A10. What work does the gas do when transitioning from state 1 to state 3 (see figure)?

Solution. Process 1–2 is isobaric: the gas pressure is equal, the volume increases by , and the gas does work. Process 2–3 is isochoric: the gas does no work. As a result, when transitioning from state 1 to state 3, the gas does 10 kJ of work.

Correct answer: 1.

A11. In a heat engine, the temperature of the heater is 600 K, the temperature of the refrigerator is 200 K less than that of the heater. The maximum possible efficiency of the machine is

Solution. The maximum possible efficiency of a heat engine is equal to the efficiency of a Carnot machine:

.

Correct answer: 4.

A12. The vessel contains constant quantity ideal gas. How will the temperature of the gas change if it goes from state 1 to state 2 (see figure)?

Solution. According to the equation of state of an ideal gas at a constant amount of gas

Correct answer: 1.

A13. The distance between two point electric charges was reduced by 3 times, and one of the charges was increased by 3 times. The forces of interaction between them

Solution. When the distance between two point electric charges decreases by 3 times, the force of interaction between them increases by 9 times. Increasing one of the charges by 3 times leads to the same increase in force. As a result, the strength of their interaction became 27 times greater.

Correct answer: 4.

A14. What will be the resistance of the circuit section (see figure) if key K is closed? (Each of the resistors has a resistance R.)

Solution. After closing the key, the terminals will be short-circuited, the resistance of this section of the circuit will become equal to zero.

Correct answer: 4.

A15. The figure shows a coil of wire through which flows electricity in the direction indicated by the arrow. The coil is located in a vertical plane. At the center of the coil is the induction vector magnetic field current is directed

Solution. According to the rule right hand: “If you clasp the solenoid (coil with current) with the palm of your right hand so that four fingers are directed along the current in the coils, then the left thumb will show the direction of the magnetic field lines inside the solenoid (coil with current).” Having mentally done specified actions, we find that at the center of the coil the magnetic field induction vector is directed horizontally to the right.

Correct answer: 3.

A16. The figure shows a graph harmonic vibrations current in the oscillatory circuit. If the coil in this circuit is replaced by another coil, the inductance of which is 4 times less, then the oscillation period will become equal to

Solution. The graph shows that the period of current oscillations in the oscillatory circuit is 4 μs. When the inductance of the coil is reduced by 4 times, the period will decrease by 2 times. After replacing the coil it will become equal to 2 µs.

Correct answer: 2.

A17. Light source S is reflected in flat mirror ab. The image S of this source in the mirror is shown in the figure

Solution. The image of an object obtained using a plane mirror is located symmetrically to the object relative to the plane of the mirror. The image of the source S in the mirror is shown in Figure 3.

Correct answer: 3.

A18. In a certain spectral range, the angle of refraction of rays at the air-glass interface decreases with increasing radiation frequency. The path of rays for the three primary colors when white light falls from air onto the interface is shown in the figure. Numbers correspond to colors

Solution. Due to the dispersion of light when passing from air to glass, the shorter its wavelength, the more the beam deviates from its original direction. U of blue color the shortest wavelength, red has the longest. The blue beam will deviate the most (1 - blue), the red beam will deviate the least (3 - red), leaving 2 - green.

Correct answer: 4.

A19. At the entrance to the electrical circuit of the apartment there is a fuse that opens the circuit at a current strength of 10 A. The voltage supplied to the circuit is 110 V. What is the maximum number of electric kettles, the power of each of which is 400 W, can be turned on simultaneously in the apartment?

Solution. An electric current with a force of 400 W passes through each kettle: 110 V 3.64 A. When two kettles are turned on, the total current strength (2 3.64 A = 7.28 A) will be less than 10 A, and when three kettles are turned on - more 10 A (3 3.64 A = 10.92 A). No more than two kettles can be turned on at the same time.

Correct answer: 2.

A20. The figure shows diagrams of four atoms corresponding to the Rutherford model of the atom. Black dots indicate electrons. The atom corresponds to the diagram

1)
2)
3)
4)

Solution. The number of electrons in a neutral atom coincides with the number of protons, which is written below before the name of the element. There are 4 electrons in an atom.

Correct answer: 1.

A21. The half-life of the nuclei of radium atoms is 1620 years. This means that in a sample containing big number radium atoms,

Solution. It is true that half of the original radium nuclei decays in 1620 years.

Correct answer: 3.

A22. Radioactive lead, having undergone one α-decay and two β-decays, turned into an isotope

Solution. During α decay, the mass of the nucleus decreases by 4 a. e.m., and during β-decay the mass does not change. After one α-decay and two β-decays, the mass of the nucleus will decrease by 4 a. eat.

During α-decay, the charge of the nucleus decreases by 2 elementary charges, and during β-decay, the charge increases by 1 elementary charge. After one α-decay and two β-decays, the charge of the nucleus will not change.

As a result, it will turn into an isotope of lead.

Correct answer: 3.

A23. The photoelectric effect is observed by illuminating the metal surface with light of a fixed frequency. In this case, the retarding potential difference is equal to U. After changing the frequency of light, the retarding potential difference increased by Δ U= 1.2 V. How much has the frequency of the incident light changed?

1)
2)
3)
4)

Solution. Let's write Einstein's equation for the photoelectric effect for the initial frequency of light and for the changed frequency. Subtracting the first from the second equality, we obtain the relation:

Correct answer: 2.

A24. The conductors are made of the same material. Which pair of conductors should be chosen in order to experimentally discover the dependence of the wire resistance on its diameter?

1)
2)
3)
4)

Solution. To experimentally discover the dependence of wire resistance on its diameter, you need to take a pair of conductors that differ only thick. The length of the conductors must be the same. You need to take a third pair of conductors.

Correct answer: 3.

A25. The dependence of the voltage on the plates of an air capacitor on the charge of this capacitor was studied. The measurement results are presented in the table.

Measurement errors q And U were equal to 0.05 µC and 0.25 kV, respectively. The capacitance of the capacitor is approximately equal to

Solution. Let us calculate the value of the capacitor capacitance () for each measurement and average the resulting values.

The calculated capacity value is closest to the third answer option.

Correct answer: 3.

Part B

IN 1. Load weight m, suspended on a spring, performs harmonic oscillations with a period T and amplitude. What will happen to the maximum potential energy of the spring, the period and frequency of oscillations, if the mass of the load is reduced at a constant amplitude?

For each position in the first column, select the corresponding position in the second and write down the selected numbers in the table under the corresponding letters.

Transfer the resulting sequence of numbers to the answer form (without spaces).

Solution. The period of oscillation is related to the mass of the load and the spring stiffness k ratio

As the mass decreases, the oscillation period will decrease (A - 2). The frequency is inversely proportional to the period, which means the frequency will increase (B - 1). Maximum potential energy spring is equal to, with a constant amplitude of oscillations it will not change (B - 3).

Answer: 213.

AT 2. Using the first law of thermodynamics, establish a correspondence between the features of the isoprocess in an ideal gas described in the first column and its name.

Transfer the resulting sequence of numbers to the answer form (without spaces or any symbols).

Solution. Internal energy ideal gas remains unchanged at a constant gas temperature, that is, in isothermal process(A - 1). There is no heat exchange with surrounding bodies in the adiabatic process (B - 4).

AT 3. A flying projectile breaks into two fragments. With respect to the direction of movement of the projectile, the first fragment flies at an angle of 90° with a speed of 50 m/s, and the second at an angle of 30° with a speed of 100 m/s. Find the ratio of the mass of the first fragment to the mass of the second fragment.

R decision. Let us depict the directions of movement of the projectile and two fragments (see figure). Let us write down the law of conservation of the projection of momentum onto an axis perpendicular to the direction of motion of the projectile:

AT 4. In a heat-insulated vessel with big amount ice at temperature is poured m= 1 kg of water at temperature . What is the mass of ice Δ m will melt when installed thermal equilibrium in a vessel? Express your answer in grams.

Solution. When cooling, the water will give up an amount of heat. This heat will melt the ice mass

Answer: 560.

AT 5. An object 6 cm high is located on the main optical axis of a thin converging lens at a distance of 30 cm from its optical center. The optical power of the lens is 5 diopters. Find the height of the object's image. Express your answer in centimeters (cm).

Solution. Let us denote the height of the object h= 6 cm, distance from lens to object, optical power lenses D= 5 diopters Using the formula for a thin lens, we determine the position of the image of the object:

.

The increase will be

.

The image height is

Part C

C1. A man with glasses walked into a warm room from the street and discovered that his glasses had fogged up. What must be the outside temperature for this phenomenon to occur? The air temperature in the room is 22 °C, and relative humidity air 50%. Explain how you got the answer.

(When answering this question, use the table for pressure saturated vapors water.)

Saturated vapor pressure of water at different temperatures

Solution. From the table we find that the saturated vapor pressure in the room is 2.64 kPa. Since the relative humidity is 50%, the partial pressure of water vapor in the room is 2.164 kPa50% = 1.32 kPa.

The first moment a person enters from the street, his glasses are at street temperature. Room air, in contact with the glasses, cools. The table shows that if the room air cools to 11 ° C or lower, when the partial pressure of water vapor becomes greater than the pressure of saturated vapor, the water vapor condenses - the glasses will fog up. The temperature outside should be no higher than 11 °C.

Answer: no higher than 11 °C.

C2. A small puck, after being hit, slides up the inclined plane from point A(see picture). At the point IN the inclined plane without a break passes into the outer surface of a horizontal pipe with a radius R. If at the point A the speed of the puck exceeds , then at the point IN the washer comes off the support. Inclined plane length AB = L= 1 m, angle α = 30°. The coefficient of friction between the inclined plane and the washer is μ = 0.2. Find the outer radius of the pipe R.

Solution. Let's find the speed of the puck at the point B using the law of conservation of energy. The change in the total mechanical energy of the washer is equal to the work of the friction force:

The separation condition is that the support reaction force is equal to zero. Centripetal acceleration caused only by gravity, while for minimal initial speed, for which separation of the washer is observed, the radius of curvature of the trajectory at the point B equals R(for higher speeds the radius will be larger):

Answer: 0.3 m.

C3. Balloon, the shell of which has mass M= 145 kg and volume, filled with hot air at normal atmospheric pressure and ambient temperature. Which minimum temperature t must there be air inside the shell for the ball to begin to rise? The shell of the ball is inextensible and has a small hole in the lower part.

Solution. The ball will begin to rise when Archimedes' force exceeds the force of gravity. Archimedes' force is . The density of outside air is

Where p- normal Atmosphere pressure, μ - molar mass of air, R- gas constant, - outside air temperature.

The mass of the ball consists of the mass of the shell and the mass of air inside the shell. The force of gravity is

Where T- temperature of the air inside the shell.

Solving the inequality, we find the minimum temperature T:

The minimum temperature of the air inside the enclosure must be 539 K or 266 °C.

Answer: 266 °C.

C4. A thin aluminum block of rectangular cross-section, having a length L= 0.5 m, slides from rest along a smooth inclined dielectric plane in a vertical magnetic field with induction B= 0.1 T (see figure). The plane is inclined to the horizontal at an angle α = 30°. The longitudinal axis of the block maintains a horizontal direction when moving. Find the magnitude of the induced emf at the ends of the block at the moment when the block passes a distance along the inclined plane l= 1.6 m.

Solution. Let's find the speed of the block in the lower position using the law of conservation of energy:

Aluminum is a conductor, so in the bar there will be induced emf. The induced emf at the ends of the bar will be equal to

Answer: 0.17 V.

C5. IN electrical circuit shown in the figure, the emf of the current source is 12 V, the capacitance of the capacitor is 2 mF, the inductance of the coil is 5 mH, the lamp resistance is 5 ohms and the resistor is 3 ohms. IN starting moment time key K is closed. What energy will be released in the lamp after the key is opened? Neglect the internal resistance of the current source, as well as the resistance of the coil and wires.

Solution. Let us introduce the following notation: ε - EMF of the current source, C- capacitor capacity, L- coil inductance, r- lamp resistance, R- resistor resistance.

While the key is closed, no current flows through the capacitor and the lamp, but current flows through the resistor and coil

The energy of the system capacitor - lamp - coil - resistor is equal to

.

After the switch is opened, transient processes will occur in the system until the capacitor is discharged and the current becomes zero. All energy will be released as heat in the lamp and resistor. At each moment of time, an amount of heat is released in the lamp, and in the resistor -. Since the same current will flow through the lamp and the resistor, the ratio of the heat generated will be in proportion to the resistances. Thus, energy will be released in the lamp

Answer: 0.115 J.

C6.-meson mass decays into two γ-quanta. Find the modulus of the momentum of one of the resulting γ-quanta in the frame of reference where the primary -meson is at rest.

Solution. In the reference frame where the primary -meson is at rest, its momentum is zero and its energy is equal to the rest energy. According to the law of conservation of momentum, γ quanta will scatter into opposite directions with the same impulses. This means that the energies of the γ-quanta are the same and, therefore, equal to half the energy of the -meson: . Then the momentum of the γ-quantum is equal to

VERY SIMPLE TASK, HELP! When moving on a horizontal surface, a sliding friction force of 50 N acts on a body weighing 10 kg. What will the sliding friction force become after reducing the body mass by 5 times, if the friction coefficient does not change?

Answers:

friction is 5 times less, which means it becomes 10 Newton

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• Find the adverbs and participial phrases While creating his wonderful canvases, Konstantin Fedorovich Yuon admires himself and makes him admire the enchanting beauty of frost-covered trees and snow-covered plains sparkling in the sun. Looking at Yuog's paintings, we remember the fabulous Russian winter with its fluffy snow, each year covering the earth with a thick cover. I also remember the light frosty haze that envelops all objects on clear, cold days and cheerful flocks of playful boys enjoying the snowy expanse. The master of the Russian winter landscape manages to achieve a silver-gray, pearly color that perfectly conveys the state of frosty, winter day. The artist said about his work: “I wanted to paint pictures the way songs are written about life, about the history of the Russian people, about nature, about ancient Russian cities.” The entire work of K. F. Yuon is a hymn to beauty, Russian life, beauty native nature, its life-affirming power.

EXERCISE:

The internal energy of a coin increases if it

1) make it rotate;

2) make you move at a higher speed;

3) throw up;

4) warm up.

SOLUTION:

Internal energy is the sum of the energies of interactions and thermal motions of molecules. It does not include the kinetic energy of the body as a whole and its energy in external fields, such as gravitational. Thus, the only way increase internal energy coins listed is to heat it.

ANSWER: 4.

A3

EXERCISE:

A stone weighing 200 g is thrown at an angle of 45 o to the horizontal with an initial speed V = 15 m/s. The modulus of gravity at the moment of the throw is equal to:

1) 0;

2) 1.33 N;

3) 3.0 N;

4) 2.0 N.

SOLUTION:

A fairly typical task for the Unified State Exam in physics with a lot of unnecessary data. The modulus of gravity acting on the stone at any time is equal to: F = mg. And the throwing angle and speed have nothing to do with it! We convert the mass into kilograms (200 g = 0.2 kg), take into account that g = 10 m/s 2, and we get: F = 0.2 x 10 = 2.0 N.

ANSWER:

4) 2.0 N.

A21

QUESTION:

Among the examples given electromagnetic waves maximum length waves has:

1) infrared radiation Sun;
2) ultraviolet radiation Sun;
3) radiation from a y-radioactive drug;
4) radiation from the radio transmitter antenna.

ANSWER:

In order to choose the right answer, it is worth knowing that the wavelengths for each of the indicated sources are within the limits:

radio waves – 10 km – 1 mm;
infrared radiation – 1 mm – 780 nm;
visible (optical) radiation – 780–380 nm;
ultraviolet – 380–10 nm;
X-ray – 10 nm – 5 pm;
gamma - less than 5 pm.

Thus, the radiation from the radio transmitter antenna has the maximum wavelength - answer: 4.

TASK

The body moves uniformly along the plane. The pressure force of the body on the plane is 20 N, the friction force is 5 N. The sliding friction coefficient is equal to:

1) 0,8;
2) 0,25;
3) 0,75;
4) 0,2.

SOLUTION

The friction force is determined by the formula: Ftr = k * N, where k is the coefficient of sliding friction, N is the pressure force of the body on the plane.

Substituting known data into this formula, we obtain the equation: 5 = k * 20; solving this equation for k, we obtain that k = 0.25. Thus, the correct answer is: 2).

A2

TASK

A piece of ice floating in a glass of fresh water, transferred to a glass of salt water. In this case, the Archimedean force acting on the ice floe is:

1) decreased, since the density fresh water less density salty;
2) decreased, as the depth of immersion of the ice in the water decreased;
3) increased, since the density of salt water is higher than the density of fresh water;
4) has not changed, since the buoyant force is equal to the weight of the ice in the air.

SOLUTION

According to Newton's first law: every body continues to be in a state of rest or uniform and rectilinear motion, while no forces act on it or their action is compensated. A piece of ice floating on the surface of water (fresh or salty) is at rest, therefore, the action of all forces on it is compensated, or, in other words, the force of gravity is equal to Archimedean force, and since the force of gravity in both cases is the same, then the Archimedean force in fresh and salt water will be the same and equal to the weight of the ice in the air.

As a result, we find that the correct answer is 4).

Atomic and nuclear physics

TASK

To accelerate spacecraft and correct their orbits, it is proposed to use solar sail– lightweight screen attached to the device large area made of a thin film that specularly reflects light. What should be the sail area S so that a device with mass m = 500 kg (including the mass of the sail) is under the influence of sun rays changed the speed by dV = 10 m/s per day? Power solar radiation is 1370 W/m2.

SOLUTION

The pressure of solar rays (light pressure) during normal incidence on a surface is expressed by the law: P = W x (1 + k) / s, where c = 3 x 10 8 m/s is the speed of light, k is the reflection coefficient. According to the condition, the surface specularly reflects light, which means k = 1, thus P = 2 x W / s. As a result, a force F = P x S will act on the sail, creating acceleration spacecraft: a = dV / dt. According to Newton’s second law, F = m x a, therefore: 2 x W x S / s = m x dV / dt, where dt is the time of action of the force - according to the condition, 1 day, or 86400 s.
Hence: S = (m x dV x s) / (2 x W x dt) = (500 x 10 x 3 x 10 8) / (2 x 1370 x 86400) = 6336 m 2.

TASK

When moving on a horizontal surface, a sliding friction force of 10 N acts on a body weighing 40 kg. What will the sliding friction force be after reducing the body's mass by 5 times, if the friction coefficient does not change?
Choose one of the options:
1) 1H;
2) 2H;
3) 4H;
4) 8H.

SOLUTION

Since the friction force Ftr = N x k, where N is the support reaction force (when moving on a horizontal surface it is equal to the force of gravity: N = m x g), k is the friction coefficient.
Thus:
Ftr = m x g x k,
This means that when the body mass decreases by 5 times, the friction force will also decrease by the same 5 times and amount to 2 N.

On this lesson, the topic of which is: “Solving problems in dynamics. Movement horizontally and along an inclined plane", we will consider solutions to a number of problems on this topic, using general algorithm solving problems in dynamics.

We continue to study the dynamics. This is a branch of physics that studies the causes of mechanical motion.

Today we will solve problems involving movement horizontally and along an inclined plane. How to solve such problems?

We have a body that is on a horizontal or inclined plane. In any case, it is subject to the force of gravity and the reaction force of the support. If the surface is not smooth, a frictional force acts on the body, directed opposite to the direction of movement. The body can be dragged by the thread, in which case the tension force of the thread will act on it. The presence of this or that force depends on the conditions of the problem, but the resultant of all forces acting on the body, in general case causes acceleration of the body. This is a consequence of Newton's second law - the main tool for solving problems in dynamics.

So, we have analyzed what happens when a body moves along a plane, determined the forces acting on the body and described the process mathematically, using Newton’s second law. This is where physics ends, and mathematics remains.

Solve equations in vector form mathematically difficult, so you need to rewrite the consequence of Newton’s second law in projections on the coordinate axes.

If the plane is inclined, it is oriented at a certain angle to the horizon, which means that the force of gravity will be directed at an angle to the plane, whether we know this angle or not. It does important choice coordinate systems.

We are free to choose, the result will not depend on the choice of coordinate system, but we need to choose one in which the mathematical transformations will be as simple as possible. We will see this in one of the problems.

And only now, when a system of equations has been obtained that describes physical process, we solve the problem mathematically: we solve equations and find the unknown.

Let's start solving problems.

A stone sliding along a horizontal surface of ice stopped after traveling a distance S = 48 m. Find the initial speed of the stone if the sliding friction force of the stone on the ice is 0.06 of the force of normal pressure of the stone on the ice.

Analysis of the condition:

The problem describes a body that moves under the influence of forces, which means we will apply Newton’s second law;

The stone is acted upon by the force of gravity, the reaction force of the support, and the force of friction. Let's mark them (see Fig. 1).

Rice. 1. Forces acting on the stone

The friction force is equal to ;

The stone stops and moves with acceleration, which, according to Newton's second law, is caused by the resultant force;

At uniformly accelerated motion the body goes through the process and gains speed.

Let's choose a coordinate system. It is convenient to direct the x-axis in the direction of movement of the stone, and the y-axis perpendicular to the x-axis (see Fig. 2).

Rice. 2. Selecting a coordinate system

Considering that the friction force is equal to , we write it in projections on the selected coordinate axes. The friction force is directed against the movement of the stone, and acceleration is also directed in the same direction (the stone slows down) (see Fig. 3):

During the stopping time, the stone according to the problem conditions will go the distance. The initial velocity is directed in the direction of the x-axis, its projection will have a “+” sign, the acceleration will be opposite the x-axis, put a “-” sign:

The body will stop, that is, its speed will be zero after time:

We have obtained a system of equations that remains to be solved and the initial speed of the stone equal to 7.6 m/s is obtained:

Now let's solve the problem of motion along an inclined plane.

A body of mass m without an initial velocity slides down an inclined plane with an angle from a height h (see Fig. 4).

Rice. 4. Drawing for problem 2

The coefficient of friction of the body on the surface is equal to . How long will it take for the body to reach the foot?

Condition Analysis

Set right triangle, in which one side and an angle are known. This means that all sides are known, and the path that the body takes is determined.

The body is acted upon by gravity, ground reaction force and friction force (see Fig. 5).

Rice. 5. Forces that act on the body

The resultant of these forces creates acceleration - we will apply Newton's second law.

In the problem, you need to find the time of motion of a body that moves with acceleration; uniformly accelerated motion is described by kinematics equations.

Let's choose a coordinate system. There is a peculiarity here: the movement of the block occurs along an inclined plane, the friction force is directed opposite to the direction of movement, the support reaction force is perpendicular to the plane, and the force of gravity is directed at an angle to the plane. It is especially important for us to choose a convenient coordinate system. For mathematical calculations, it is convenient to direct the coordinate axes as shown in the figure: the x axis is along the direction of movement of the block, the y axis is perpendicular to the surface (see Fig. 6).

Rice. 6. Selecting a coordinate system

Let's apply Newton's second law:

Considering that the friction force is equal to , we write it in projections on the selected coordinate axes.

The force of gravity is directed at an angle to both coordinate axes. Triangles ABC and ABC are similar, and the angle equal to angle cab. Consequently, the projection of gravity on the x-axis is equal to, and on the y-axis - (see Fig. 7).

Rice. 7. Projections of forces on the coordinate axes

Finding gravity projections

The body travels a path AB equal to triangle ABC. The path traveled by a body during uniformly accelerated motion without initial speed is equal to:

We have obtained a system of equations from which it remains to find the time:

Mathematical part of solving the problem

Selecting a coordinate system

A block with a thread attached rests on an inclined plane with an inclination angle of 30 0. At what minimum strength tension of the thread, will the block move if you pull the thread down so that it is parallel to the plane? The mass of the block is 0.5 kg, the sliding friction coefficient of the block on the plane is 0.7, the acceleration free fall take equal to 10 m/s 2.

Condition Analysis

The problem describes a body that is subject to the force of gravity, the reaction force of the support, the force of friction and the tension force of the thread (see Fig. 12).

Rice. 12. Action of forces on the body

The body is pulled down, the friction force is directed against the possible direction of movement.

According to the conditions of the problem, at a certain minimum value of the tension force of the thread, the block moves from its place, the block will not accelerate, the acceleration is zero. We will apply Newton's second law, the acceleration is 0.

Let's choose a coordinate system. We have already seen from the example previous task, which is convenient to direct the x-axis parallel to the plane (see Fig. 13), and the y-axis perpendicular to the plane.

Rice. 13. Selecting a coordinate system

According to Newton's second law, the sum of forces acting on the block is equal, in our case:

Considering that the friction force is equal to , we write in projections onto the selected coordinate axes:

We have obtained a system of equations, solving which we find minimum value.

Mathematical part of solving the problem

As you can see, problems involving the motion of bodies along an inclined plane, like most other problems in dynamics, come down to the application of Newton's laws in a selected convenient coordinate system.

This concludes our lesson, thank you for your attention!

Bibliography

1. Sokolovich Yu.A., Bogdanova G.S. Physics: A reference book with examples of problem solving. - 2nd edition repartition. - X.: Vesta: Ranok Publishing House, 2005. - 464 p.
2. A.V. Rusakov, V.G. Sukhov. Collection of problems in physics (physics and mathematics school No. 2, Sergiev Posad). - 1998

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