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IZhO 2024, Теория — Задача 1

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XX International Zhautykov Olympiad/Theoretical Competition · 9 януари 2024 г. · 10 т.

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THEORETICAL COMPETITION

January 9, 2024

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The time available for the theoretical competition is 4 hours. There are three questions.
Use only the pen provided.
You can use your own calculator for numerical calculations. If you don't have one, please ask for it from Olympiad organizers.
You are provided with Writing sheet and additional paper. You can use the additional paper for drafts of your solutions but these papers will not be checked. Your final solutions which will be evaluated should be on the Writing sheets. Please use as little text as possible. You should mostly use equations, numbers, figures and plots.
Use only the front side of Writing sheets. Write only inside the boxed area.
Begin each question on a separate sheet of paper.
Fill the boxes at the top of each sheet of paper with your country (Country), your student code (Student Code), the question number (Question Number), the progressive number of each sheet (Page Number), and the total number of Writing sheets used (Total Number of Pages). If you use some blank Writing sheets for notes that you do not wish to be evaluated, put a large X across the entire sheet and do not include it in your numbering.
At the end of the exam, arrange all sheets for each problem in the following order:
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Place the papers inside the envelope and leave everything on your desk. You are not allowed to take any paper out of the room.

Attention

SOLUTIONS TO THE PROBLEMS OF THE THEORETICAL COMPETITION

Attention. Points in grading are not divided!

Условие

This problem consists of three independent parts.

Two views of the bicone on V-shaped slats. Left (top view): a bicone of height 2h sitting across two slats that converge at the vertex O with dihedral angle 2γ; the x axis runs from O towards the bicone; g is directed into the page (⊗ g). Right (side view): the bicone base circle of diameter 2R resting on a slat, with the vertex O on the right and the vector g pointing down.

Thermally insulated vessel divided by a partition into two halves; left half labelled V0, P0, T1 and N2, right half labelled V0, P0, T2 and N2.

1.1 On two identical horizontal V-shaped slats (see figure below), forming a dihedral angle of $2\gamma$ ( $\tan\gamma = 0.100$ ), there is a solid bicone with a base diameter of $D = 2R = 8.00\ \mathrm{cm}$ and a distance between the vertices of $2h = 20.0\ \mathrm{cm}$ , representing two identical homogeneous cones, rigidly fastened by their bases. First, the bicone is held so that its center of mass is right above the vertex of the dihedral angle, symmetrically relative to its bisectoral plane. Then it is released and it begins to roll along the slats without slipping. Find the dependence of the bicone speed of translational motion of the $v(x)$ on the $x$ coordinate measured horizontally from the vertex of the dihedral angle to the center of mass of the bicone. Calculate the speed of the bicone $v_0$ at $x_0 = 50.0\ \mathrm{cm}$ . Determine the maximum angular velocity $\omega_{max}$ , that the bicone is to reach during its motion. Consider that the height of the slats exceeds the radius of the cone bases, and the acceleration of gravity is $g = 9.80\ \mathrm{m/c^2}$ . [3 т.]

1.2 A thermally insulated vessel with a volume of $2V_0 = 2.00\ \mathrm{l}$ is divided by a thin weightless movable heat-conducting partition into two equal parts, each containing nitrogen $N_2$ at a pressure $P_0 = 100\ \mathrm{kPa}$ . The initial temperature of nitrogen in one half of the vessel is $T_1 = 240\ \mathrm{K}$ , and it is $T_2 = 360\ \mathrm{K}$ in the other. As a result of heat exchange, the partition begins to move slowly without friction. 1) Determine the amount of heat $Q$ that will be exchanged between parts of the vessel by the time complete thermodynamic equilibrium is established; 2) Calculate the minimum work $A'$ that must be done to return the partition to its original position. Neglect the heat capacities of the vessel and partition. [4 т.]

1.3 An electric motor connected to a battery lifts a load of mass $m$ at a constant speed to a certain height $h$ in time $t_1 = 1.00\ \mathrm{min}$ . In this case, the load is tied to a thread that is wound on a uniformly rotating electric motor shaft. The time it takes to lift two loads with a total mass of $2m$ to the same height turns out to be $\Delta t = 7.00\ \mathrm{s}$ longer. 1) Find the maximum number $n_{max}$ of loads of the same mass $m$ that such an electric motor can lift; 2) Calculate the time $t_{max}$ that the electric motor is to require to uniformly lift the maximum number of loads $n_{max}$ to the height $h$ . Consider that the torque of an electric motor is proportional to the current flowing through its winding. [3 т.]

Отговори

Решение

Problem 1.1 (3.0 points)

Since the bicone rolls along the slats without slipping, its translational speed $v$ and angular speed of rotation $\omega$ are related through the rolling radius $r$ by the relation

v = \omega r. \qquad (1)

The kinetic energy of the translational motion of the bicone is equal to

W_k = \frac{mv^2}{2}, \qquad (2)

and the corresponding rotational energy is written as

W_r = \frac{I\omega^2}{2}, \qquad (3)

where the moment of inertia of the bicone is introduced

I = \frac{3}{10} mR^2. \qquad (4)

The change in potential energy of the bicone during its motion is

W_p = -mgR\left(1 - \frac{r}{R}\right), \qquad (5)

and according to the law of conservation of energy the relation must be satisfied

W_k + W_r = -W_p. \qquad (6)

From the geometric relationships the relation between the rolling radius $r$ and the coordinate $x$ is obtained in the following form

r = R\left(1 - \frac{x}{h}\tan\gamma\right), \qquad (7)

so, putting together equations (1)-(7), we get

v(x) = \sqrt{gD\,\frac{\frac{x}{h}\tan\gamma}{1 + 3/10\left(1 - \frac{x}{h}\tan\gamma\right)^2}}. \qquad (8)

In particular, for the value $x_0 = 50.0\ \mathrm{cm}$ the calculations

v_0 = v(x_0) = 42.2\ \mathrm{cm/s}. \qquad (9)

From the same expressions (1)-(7) the dependence of the square of the angular velocity of rotation on the rolling radius is found as

\omega^2 = \frac{2g}{R}\cdot\frac{mR^2}{I}\cdot\frac{\left(1 - \frac{r}{R}\right)}{\left(1 + \frac{mr^2}{I}\right)}, \qquad (10)

which has a maximum value at $r = 0$ , equal to

\omega_{max} = \sqrt{\frac{20g}{3R}} = 40.4\ \mathrm{rad/s}. \qquad (11)

It is interesting to note that the bicone in this position actually rotates in one place, that is, its translational speed actually becomes zero in accordance with formula (1).

Content	Points
Formula (1): $v = \omega r$	0.2
Formula (2): $W_k = \frac{mv^2}{2}$	0.2
Formula (3): $W_r = \frac{I\omega^2}{2}$	0.2
Formula (4): $I = \frac{3}{10} mR^2$	0.5
Formula (5): $W_p = -mgR\left(1 - \frac{r}{R}\right)$	0.2
Formula (6): $W_k + W_r = -W_p$	0.2
Formula (7): $r = R\left(1 - \frac{x}{h}\tan\gamma\right)$	0.5
Formula (8): $v(x) = \sqrt{gD\,\frac{\frac{x}{h}\tan\gamma}{1 + 0{,}3/\left(1 - \frac{x}{h}\tan\gamma\right)^2}}$	0.2
Formula (9): $v_0 = 42.2\ \mathrm{cm/s}$	0.2
Formula (10): $\omega^2 = \frac{2g}{R}\cdot\frac{mR^2}{I}\cdot\frac{\left(1 - \frac{r}{R}\right)}{\left(1 + \frac{mr^2}{I}\right)}$	0.2
Formula (11): $\omega_{max} = \sqrt{\frac{20g}{3R}}$	0.2
Numerical value in formula (11): $\omega_{max} = 40.4\ \mathrm{rad/s}$	0.2
Total	3.0

Problem 1.2 (4.0 points)

The total internal energy of the system as a whole does not change during the process of temperature equalization, since it does not do any external work and no heat is supplied to it, that is

U = U_0. \qquad (1)

It follows that when the partition moves, the gas pressures in each part of the vessel, which are equal to each other in this quasi-static process, do not change. Indeed, the initial internal energy of the system is equal to

U_0 = \frac{C_V}{R} P_0 2V_0, \qquad (2)

where

C_V = \frac{5}{2} R. \qquad (3)

The internal energy at an arbitrary moment of time is

U = \frac{C_V}{R} P(2V_0), \qquad (4)

where $P$ stands for the gas pressure in both parts of the vessel, which is obtained from equations (1)-(4)

P = P_0, \qquad (5)

that is, the processes occurring with gases are isobaric.

Let us write down the equation of state of an ideal gas for each part of the vessel at the initial moment of time as

P_0 V_0 = \nu_1 R T_1, \qquad (6)

P_0 V_0 = \nu_2 R T_2, \qquad (7)

where $\nu_1$ and $\nu_2$ denote the number of moles of nitrogen in each half of the vessel, respectively.

In the final state, the gas is at a certain temperature $T_0$ and its equation of state has the form

P_0 2V_0 = \nu R T_0, \qquad (8)

where the total number of moles of nitrogen in the vessel is equal to

\nu = \nu_1 + \nu_2. \qquad (9)

From formulas (6)-(9) we determine the final temperature of the gas in the vessel

T_0 = \frac{2T_1T_2}{T_1 + T_2}. \qquad (10)

Since the process is isobaric, the amount of heat $Q$ exchanged between the two parts of the vessel is derived as

Q = C_P \nu_1 (T_0 - T_1) = \frac{7}{2} P_0 V_0 \cdot \frac{T_2 - T_1}{T_1 + T_2} = 70.0\ \text{Дж}, \qquad (11)

where

C_p = C_V + R. \qquad (12)

When the partition returns to its original position, the work of external forces is to be minimal if the movement process is slow, quasi-equilibrium without disturbing the thermal equilibrium between both parts of the vessel

T_1' = T_2' = T, \qquad (13)

that is, unlike the previous case, in both parts of the vessel the pressure is not the same, but the temperature is, which, however, changes.

The initial volumes of each part of the vessel are determined by the Gay-Lussac law and are equal to

V_{01} = \frac{V_0 T_0}{T_1}, \qquad (14)

V_{02} = \frac{V_0 T_0}{T_2}. \qquad (15)

Let us denote the pressure in parts of the vessel as $P_1$ and $P_2$ , and the corresponding volumes as $V_1$ and $V_2$ . In accordance with the equation of state, at an arbitrary moment of time the following relations must be satisfied

P_1 V_1 = \nu_1 R T, \qquad (16)

P_2 V_2 = \nu_2 R T. \qquad (17)

In this process, the change in the internal energy of the gas in the system is equal to

dU = \nu C_V dT, \qquad (18)

and if the gas does work $\delta A$ on external bodies, then according to the first law of thermodynamics, under the conditions of vessel thermal insulation as a whole, the supplied amount of heat becomes zero

\delta Q = dU + \delta A = 0. \qquad (19)

In a quasi-static process, the work of the gas as a whole consists of the work of the gases in each part of the vessel

\delta A = P_1 dV_1 + P_2 dV_2. \qquad (20)

Writing (18)-(20) together and using equations (3), (6), (7), (9), (10), (14) and (15)-(17), we obtain

\frac{5}{T_0} dT + \frac{T}{T_1 V_1} dV_1 + \frac{T}{T_2 V_2} dV_2 = 0, \qquad (21)

whose integration gives the final temperature

T_f = T_0 \left(\frac{T_0}{T_1}\right)^{\frac{T_0}{5T_1}} \left(\frac{T_0}{T_2}\right)^{\frac{T_0}{5T_2}} = 290\ \mathrm{K}. \qquad (22)

The work done by external forces on the partition to move it is opposite in sign to the work done by the gas itself, therefore, from expressions (16) and (17) the work can easily be found in the form

A' = -A = \Delta U = \nu C_V (T_f - T_0) = 5P_0V_0\,\frac{T_f - T_0}{T_0} = 4.04\ \mathrm{J}. \qquad (23)

Alternative solution. The process of returning the partition to its original position is adiabatic, i.e. occurs without change in entropy

S = const. \qquad (24)

The total change in entropy of an ideal gas in both parts of the vessel is equal to:

\Delta S = \nu_1 C_V \ln\frac{T_f}{T_0} + \nu_1 R \ln\frac{V_0}{V_{01}} + \nu_2 C_V \ln\frac{T_f}{T_0} + \nu_2 R \ln\frac{V_0}{V_{02}} = 0, \qquad (25)

from where, using (6), (7), (10), (14), (15) we obtain the final temperature of the system

T_f = T_0 \left(\frac{T_0}{T_1}\right)^{\frac{T_0}{5T_1}} \left(\frac{T_0}{T_2}\right)^{\frac{T_0}{5T_2}} = 290\ \mathrm{K}. \qquad (26)

The work of the external force is spent on changing the internal energy of the gas:

A' = \Delta U = (\nu_1 + \nu_2) C_V (T_f - T_0) = 5P_0V_0\,\frac{T_f - T_0}{T_0} = 4.04\ \mathrm{J}. \qquad (27)

Content	Points
Formula (1): $U = U_0$	0.1
Formula (2): $U_0 = \frac{C_V}{R} P_0 2V_0$	0.1
Formula (3): $C_V = \frac{5}{2} R$	0.1
Formula (4): $U = \frac{C_V}{R} P(2V_0)$	0.1
Formula (5): $P = P_0$	0.4
Formula (6): $P_0 V_0 = \nu_1 R T_1$	0.1
Formula (7): $P_0 V_0 = \nu_2 R T_2$	0.1
Formula (8): $P_0 2V_0 = \nu R T_0$	0.1
Formula (9): $\nu = \nu_1 + \nu_2$	0.1
Formula (10): $T_0 = \frac{2T_1T_2}{T_1 + T_2}$	0.4
Formula (11): $Q = \frac{7}{2} P_0 V_0 \cdot \frac{T_2 - T_1}{T_1 + T_2}$	0.4
Numerical value in formula (11): $Q = 70.0\ \mathrm{J}$	0.1
Formula (12): $C_p = C_V + R$	0.1
Formula (13): $T = T_1 = T_2$	0.4
Formula (14): $V_{01} = \frac{V_0 T_0}{T_1}$	0.1
Formula (15): $V_{02} = \frac{V_0 T_0}{T_2}$	0.1
Formula (16): $P_1 V_1 = \nu_1 R T$	0.1
Formula (17): $P_2 V_2 = \nu_2 R$	0.1
Formula (18): $dU = \nu C_V dT$	0.1
Formula (19): $\delta Q = dU + \delta A = 0$	0.1
Formula (20): $\delta A = P_1 dV_1 + P_2 dV_2$	0.1
Formula (21): $\frac{5}{T_0} dT + \frac{T}{T_1 V_1} dV_1 + \frac{T}{T_2 V_2} dV_2 = 0$	0.1
Formula (22): $T_f = T_0 \left(\frac{T_0}{T_1}\right)^{\frac{T_0}{5T_1}} \left(\frac{T_0}{T_2}\right)^{\frac{T_0}{5T_2}}$	0.2
Formula (23): $A' = \nu C_V \left(T_f - T_0\right) = 5P_0V_0\,\frac{T_f - T_0}{T_0}$	0.2
Numerical value in formula (23): $A' = 4.04\ \mathrm{J}$	0.2
Alternative solution to 2)
Formula (13): $T = T_1 = T_2$	0.4
Formula (14): $V_{01} = \frac{V_0 T_0}{T_1}$	0.1
Formula (15): $V_{02} = \frac{V_0 T_0}{T_2}$	0.1
Formula (24): $S = const.$	0.2
Formula (25): $\Delta S = \nu_1 C_V \ln\frac{T_f}{T_0} + \nu_1 R \ln\frac{V_0}{V_{01}} + \nu_2 C_V \ln\frac{T_f}{T_0} + \nu_2 R \ln\frac{V_0}{V_{02}} = 0$	0.4
Formula (26): $T_f = T_0 \left(\frac{T_0}{T_1}\right)^{\frac{T_0}{5T_1}} \left(\frac{T_0}{T_2}\right)^{\frac{T_0}{5T_2}}$	0.2
Formula (27): $A' = \nu C_V \left(T_f - T_0\right) = 5P_0V_0\,\frac{T_f - T_0}{T_0}$	0.2
Numerical value in formula (27): $A' = 4.04\ \mathrm{J}$	0.2
Total	4.0

Problems 1.3 (3.0 points)

The power of the battery is spent on mechanical work of lifting the load, as well as on the power of thermal losses on the internal resistance of the battery $r$ and on the ohmic resistance $R$ of the motor winding. If the torque required to uniformly lift a load of mass $m_1 = m$ is equal to $M$ and the shaft rotates at an angular speed $\omega_1 = \varphi/t_1$ , where $\varphi$ is the angle of rotation of the shaft when lifting the load to the height $h$ , then according to the law of conservation of energy:

UI = M\omega_1 + I^2(r + R), \qquad (1)

where $U$ is the emf of the battery, $I$ designates the current strength that creates the torque necessary to lift a load of mass $m$ .

Taking into account the proportionality of torque to current strength

M = \alpha I, \qquad (2)

with $\alpha$ being the proportionality coefficient, equation (1) is rewritten in the form

U = \alpha\omega_1 + I(r + R). \qquad (3)

Obviously, to uniformly lift a load of mass $m_2 = 2m$ , twice the torque and, accordingly, twice the current are required. If the shaft rotation speed is equal to $\omega_2 = \frac{\varphi}{t_2}$ , where $t_2 = t_1 + \Delta t$ is the time of lifting the load $m_2$ , then the corresponding equation has the form:

U = \alpha\omega_2 + 2I(r + R). \qquad (4)

For uniform lifting of a load with a mass of $m_n = nm$ , the required current is equal to $I_n = nI$ . If this load is lifted in time $t_n$ at a speed of $\omega_n = \frac{\varphi}{t_n}$ , then the corresponding equation looks like this:

U = \alpha\omega_n + nI(r + R). \qquad (5)

From equations (3)-(5) $\omega_n$ is easily expressed through $\omega_1, \omega_2$ :

\omega_n = (n - 1)\omega_2 - (n - 2)\omega_1, \qquad (6)

whence the time for lifting a load of mass $m_n$ is equal to

t_n = \frac{t_1 t_2}{(n - 1)t_1 - (n - 2)t_2} = \frac{t_1 t_2}{t_2 - (n - 1)\Delta t}. \qquad (7)

From the condition

t_n > 0, \qquad (8)

we find that $n < \frac{t_2}{\Delta t} + 1$ , which means that the number of loads $n$ should not exceed

n_{max} = \left[\frac{t_2}{\Delta t} + 1\right] = \left[\frac{t_1 + 2\Delta t}{\Delta t}\right] = [10{,}6\ \ldots] = 10. \qquad (9)

The lifting time of $n = n_{max}$ loads is equal to

t_{max} = \frac{t_1 t_2}{t_2 - (n_{max} - 1)\Delta t} = \frac{t_1(t_1 + \Delta t)}{t_1 - (n_{max} - 2)\Delta t} = 1.01\cdot 10^3\ \mathrm{s}. \qquad (10)

Content	Points
Formula (1): $UI = M\omega_1 + I^2(r + R)$	0.3
Formula (2): $M = \alpha I$	0.1
Formula (3): $U = \alpha\omega_1 + I(r + R)$	0.2
Formula (4): $U = \alpha\omega_2 + 2I(r + R)$	0.2
Formula (5): $U = \alpha\omega_n + nI(r + R)$	0.2
Formula (6): $\omega_n = (n - 1)\omega_2 - (n - 2)\omega_1$	0.2
Formula (7): $t_n = \frac{t_1 t_2}{(n - 1)t_1 - (n - 2)t_2} = \frac{t_1 t_2}{t_2 - (n - 1)\Delta t}$	0.2
Formula (8): $t_n > 0$	0.4
Formula (9): $n_{max} = \left[\frac{t_2}{\Delta t} + 1\right] = \left[\frac{t_1 + 2\Delta t}{\Delta t}\right]$	0.4
Numerical value in formula (9): $n_{max} = 10$	0,2
Formula (10): $t_{max} = \frac{t_1 t_2}{t_2 - (n_{max} - 1)\Delta t} = \frac{t_1(t_1 + \Delta t)}{t_1 - (n_{max} - 2)\Delta t}$	0,4
Numerical value in formula (10): $t_{max} = 1.01\cdot 10^3\ \mathrm{s}$	0,2
Total	3.0

Mathematical hint for the theoretical competition

Mathematical hint for the theoretical competition

The following formulas may be useful:

$\int x^n dx = \frac{x^{n+1}}{n+1} + C$ , where $n \neq -1$ is a fixed number, $C$ refers to an arbitrary constant;

$\int \frac{dx}{x} = \ln|x| + C$ , where $C$ stands for an arbitrary constant;

$(1 + x)^\gamma \approx 1 + \gamma x + \frac{\gamma(\gamma - 1)}{2}x^2$ , for $|x| \ll 1$ and any value of $\gamma$ ;

$\tan x \approx \sin x \approx x$ , for $|x| \ll 1$ ;

$\ln(1 + x) \approx x$ , for $|x| \ll 1$ .

Оригинал в Архива: IZhO-2024-Th_eng.pdf · официални решения: IZhO-2024-Theory_multi_sol.pdf