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IZhO 2021, Теория — Задача 1

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XVII International Zhautykov Olympiad/Theoretical Competition — THEORETICAL COMPETITION, January 8, 2021 · 8 януари 2021 г. · 10 т.

Условие

This problem consists of three independent parts.

1.1 Water of mass density $\rho = 1.00g/sm^3$ is poured into a vertical $U$ -shaped tube of small constant cross-section $s = 8.00sm^2$ , such that the total length of water in both legs is $l = 50.0sm$ . Two pistons are placed into one leg of the tube with the spring of stiffness $k = 1.00N/m$ in between. The pistons are watertight and can slide along the tube without friction. At the initial moment, a weight of mass $m = 10.0g$ is placed on the upper piston. Determine the possible frequencies of small harmonic vibrations of the system near a new equilibrium position. The mass of the pistons and the spring can be neglected, the acceleration of gravity is equal to $g = 9.80m/s^2$ . Consider water as an ideal incompressible liquid. [4 т.]

Vertical U-shaped tube partly filled with water (shaded blue); in the left leg two pistons with a spring k between them, a weight m on the upper piston; an arrow g points downward on the right.

1.2 Some insects, such as water striders, are capable of moving freely on the water surface, because their legs are densely covered with non-wetting hairs. To understand why this turns possible, consider the following model problem. A square plate with the side $a = 10.0sm$ and thickness $h = 2.00mm$ is carefully laid on the water surface. The density of the plate material is equal to $\rho = 1.10g/sm^3$ , the density of water is $\rho_0 = 1.00g/sm^3$ , and its surface tension constitutes $\sigma = 7.30 \cdot 10^{-2}N/m$ . Find the maximum mass $m$ of a weight that can be put on the plate so that it still does not sink. Assume that the weight does not deform the plate, the acceleration of free fall is $g = 9.80m/s^2$ . [3 т.]

1.3 The electrical circuit shown schematically in the figure consists of three capacitors with capacities $C_1$ , $C_2$ and $C_3$ , a coil of inductance $L$ and a source of constant voltage $U_0$ . At the initial moment of time, the capacitors are not charged, and the current in the coil is zero. The switch $K$ is shorted. Find the maximum current $I_{\max}$ through the coil and determine the minimum voltage $U_{\min}$ across the capacitor $C_2$ . Assume that the resistance of the connecting wires is rather small. [3 т.]

Circuit: switch K and source U0 on the left branch, capacitor C1 and coil L along the top, capacitor C2 in the middle vertical branch, capacitor C3 on the bottom right branch.

Отговори

Решение

Attention. Points in grading are not divided!

Problem 1.1 (4.0 points)

Let $x$ be the spring compression, and $y$ be the change in the water level in the tube leg in which the pistons are located. When water moves in a tube, an inertial force acts on the body, which is equal to

F = -m\ddot{y}, \qquad (1)

so that the equation of the weight motion is written as

m\ddot{x} = -kx + mg - m\ddot{y}. \qquad (2)

The equation of motion of water in the tube has the form

\rho sl\ddot{y} = -2\rho sgy + kx. \qquad (3)

The new equilibrium position is determined by the conditions $x = x_0 = const$ and $y = y_0 = const$ , so that substitution into equations (2) and (3) gives rise to

x_0 = \frac{mg}{k}, \qquad (4)

y_0 = \frac{kx_0}{2\rho gs} = \frac{m}{2\rho s}. \qquad (5)

According to the problem statement, it is said that the system performs harmonic oscillations about the new equilibrium position, therefore, a solution to equations (2) and (3) is sought in the following form

x = x_0 + A\cos\omega t, \qquad (6)

y = y_0 + B\cos\omega t, \qquad (7)

and after substitution we obtain the following set of equations

A(\omega_1^2 - \omega^2) = B\omega^2, \qquad (8)

A\omega_3^2 = B(\omega_2^2 - \omega^2), \qquad (9)

where $\omega_1^2 = \dfrac{k}{m}$ , $\omega_2^2 = \dfrac{2g}{l}$ , $\omega_3^2 = \dfrac{k}{\rho sl}$ .

After dividing equations (8) and (9), we obtain a quadratic equation for a possible oscillation frequencies

(\omega_1^2 - \omega^2)(\omega_2^2 - \omega^2) = \omega^2\omega_3^2, \qquad (10)

which admits the solution

\omega_{1,2}^2 = \frac{\omega_1^2 + \omega_2^2 + \omega_3^2 \pm \sqrt{(\omega_1^2 + \omega_2^2 + \omega_3^2)^2 - 4\omega_1^2\omega_2^2}}{2}. \qquad (11)

Note that both roots are always positive and give the following possible frequencies of harmonic oscillations

\omega_1 = \sqrt{\frac{\omega_1^2 + \omega_2^2 + \omega_3^2 - \sqrt{(\omega_1^2 + \omega_2^2 + \omega_3^2)^2 - 4\omega_1^2\omega_2^2}}{2}} = 5.19 s^{-1}, \qquad (12)

\omega_2 = \sqrt{\frac{\omega_1^2 + \omega_2^2 + \omega_3^2 + \sqrt{(\omega_1^2 + \omega_2^2 + \omega_3^2)^2 - 4\omega_1^2\omega_2^2}}{2}} = 12.07 s^{-1}. \qquad (13)

In reality, the motion of the system is represented by the addition of harmonic oscillations with frequencies (12) and (13).

Content	Points
Formula (1): $F = -m\ddot{y}$	0.3
Formula (2): $m\ddot{x} = -kx + mg - m\ddot{y}$	0.3
Formula (3): $\rho sl\ddot{y} = -2\rho sgy + kx$	0.3
Formula (4): $x_0 = \dfrac{mg}{k}$	0.3
Formula (5): $y_0 = \dfrac{kx_0}{2\rho gs} = \dfrac{m}{2\rho s}$	0.3
Formula (6): $x = x_0 + A\cos\omega t$	0.3
Formula (7): $y = y_0 + B\cos\omega t$	0.3
Formula (8): $A(\omega_1^2 - \omega^2) = B\omega^2$	0.2
Formula (9): $A\omega_3^2 = B(\omega_2^2 - \omega^2)$	0.2
Formula (10): $(\omega_1^2 - \omega^2)(\omega_2^2 - \omega^2) = \omega^2\omega_3^2$	0.4
Formula (11): $\omega_{1,2}^2 = \dfrac{\omega_1^2 + \omega_2^2 + \omega_3^2 \pm \sqrt{(\omega_1^2 + \omega_2^2 + \omega_3^2)^2 - 4\omega_1^2\omega_2^2}}{2}$	0.3
Formula (12): $\omega_1 = \sqrt{\dfrac{\omega_1^2 + \omega_2^2 + \omega_3^2 - \sqrt{(\omega_1^2 + \omega_2^2 + \omega_3^2)^2 - 4\omega_1^2\omega_2^2}}{2}}$	0.2
Numerical value in formula (12): $\omega_1 = 5.19 s^{-1}$	0.2
Formula (13): $\omega_2 = \sqrt{\dfrac{\omega_1^2 + \omega_2^2 + \omega_3^2 + \sqrt{(\omega_1^2 + \omega_2^2 + \omega_3^2)^2 - 4\omega_1^2\omega_2^2}}{2}}$	0.2
Numerical value in formula (13): $\omega_2 = 12.07 s^{-1}$	0.2
Total	4.0

Problem 1.2 (3.0 points)

The plate has volume

V = a^2h, \qquad (1)

and it is subject to the gravity force

F_p = \rho Vg. \qquad (2)

At the line of contact between the plate and water, a difference in water levels occurs, as shown in the figure below.

Edge of the plate floating below the water level: the water surface curves down by height H to the plate of thickness h; surface-tension forces F_σ act on a water strip of width l, and the pressure force F_p acts horizontally.

As a result, on the lower surface of the plate with the area

S = a^2 \qquad (3)

differential pressure applies

\Delta p = \rho_0 g(H + h), \qquad (4)

which results in a vertically upward force

F = \Delta pS. \qquad (5)

To determine the value of $H$ , we select a certain volume of water with the width $l$ near its contact line with the plate. It is subject to the surface tension force equal to

F_\sigma = 2\sigma l, \qquad (6)

as well as the force due to the pressure of the liquid column

F_{\bar{p}} = \bar{p}\Delta S, \qquad (7)

where the average pressure ia found as

\bar{p} = \frac{1}{2}\rho_0 gH \qquad (8)

together with the cross-sectional area

\Delta S = Hl. \qquad (9)

From the water equilibrium condition

F_\sigma = F_{\bar{p}} \qquad (10)

it follows that the height difference is obtained as

H = 2\sqrt{\frac{\sigma}{\rho_0 g}}. \qquad (11)

The additional weight on the plate is acted upon by the gravity force

F_m = mg, \qquad (12)

and equilibrium condition

F_p + F_m = F \qquad (13)

the mass of the weight is finally derived as

m = (\rho_0 - \rho)a^2h + 2a^2\sqrt{\frac{\sigma\rho_0}{g}} = 52.6g. \qquad (14)

Content	Points
Formula (1): $V = a^2h$	0.2
Formula (2): $F_p = \rho Vg$	0.2
Formula (3): $S = a^2$	0.2
Formula (4): $\Delta p = \rho_0 g(H + h)$	0.2
Formula (5): $F = \Delta pS$	0.2
Formula (6): $F_\sigma = 2\sigma l$	0.2
Formula (7): $F_{\bar{p}} = \bar{p}\Delta S$	0.2
Formula (8): $\bar{p} = \dfrac{1}{2}\rho_0 gH$	0.2
Formula (9): $\Delta S = Hl$	0.2
Formula (10): $F_\sigma = F_{\bar{p}}$	0.2
Formula (11): $H = 2\sqrt{\dfrac{\sigma}{\rho_0 g}}$	0.2
Formula (12): $F_m = mg$	0.2
Formula (13): $F_p + F_m = F$	0.2
Formula (14): $m = (\rho_0 - \rho)a^2h + 2a^2\sqrt{\dfrac{\sigma\rho_0}{g}}$	0.2
Numerical value in formula (14): $m = 52.6g$	0.2
Total	3.0

Problem 1.3 (3.0 points)

The current through the coil cannot change instantly and immediately after the key $K$ is shorted it remains equal to zero. At the same time, since the resistance of the connecting wires is very small, the capacitors $C_1$ and $C_2$ are almost instantly charged up to charges $q_{10}$ and $q_{20}$ respectively, whereas the capacitor $C_3$ remains uncharged

q_{30} = 0, \qquad (1)

since it can only be charged through the coil. Note that Joule heat is generated in the connecting wires.

Thus, at the initial moment of time, the capacitors $C_1$ and $C_2$ are connected in series to a constant voltage source $U_0$ and their charges are equal

q_{10} = q_{20}, \qquad (2)

and the corresponding voltages add up, so that

\frac{q_{10}}{C_1} + \frac{q_{20}}{C_2} = U_0. \qquad (3)

Thus, we find from equations (2) and (3) that

q_{10} = q_{20} = \frac{C_1C_2}{C_1 + C_2}U_0. \qquad (4)

The total energy of the system immediately after the key $K$ shortening turns out to be

W_0 = \frac{q_{10}^2}{2C_1} + \frac{q_{20}^2}{2C_2} = \frac{C_1C_2U_0^2}{2(C_1 + C_2)}. \qquad (5)

After charging the capacitors $C_1$ and $C_2$ , the current through the coil starts to increase and harmonic oscillations are generated in the system, at which Joule losses can already be neglected, since the resistance of the connecting wires is very small.

Note that at that moment in time when the current in the coil is maximum, the voltage across it is zero and the capacitors $C_2$ and $C_3$ turn out to be connected in parallel. For such a connection of capacitors, the following relations for charges are satisfied

q_1 = q_2 + q_3. \qquad (6)

\frac{q_2}{C_2} = \frac{q_3}{C_3}. \qquad (7)

\frac{q_1}{C_1} + \frac{q_2}{C_2} = U_0. \qquad (8)

Solving together the set of equations (6)-(8), we find the charges of the capacitors

q_1 = \frac{C_1(C_2 + C_3)}{C_1 + C_2 + C_3}U_0. \qquad (9)

q_2 = \frac{C_1C_2}{C_1 + C_2 + C_3}U_0. \qquad (10)

q_3 = \frac{C_1C_3}{C_1 + C_2 + C_3}U_0, \qquad (11)

and the energy of the system in this state is obviously equal to

W = \frac{C_1(C_2 + C_3)U_0^2}{2(C_1 + C_2 + C_3)} + \frac{LI_{\max}^2}{2}. \qquad (12)

In this case, the work of the source is found as

A = (q_1 - q_{10})U_0, \qquad (13)

and the energy conservation law is written in the following form

W_0 + A = W, \qquad (14)

which provides the maximum current

I_{\max} = \sqrt{\frac{C_3}{(C_1 + C_2)(C_1 + C_2 + C_3)L}}C_1U_0. \qquad (15)

Finding the minimum voltage $U_{\min}$ across the capacitor $C_2$ is a slightly more difficult task that has a rather simple solution. It is obvious that harmonic oscillations occur in the system, at which the potential energy is constantly transformed into kinetic energy and backwards. For the presented electrical circuit, the role of the kinetic energy is played by the energy of the coil. Therefore, when the current through the coil is zero, then the system is in its large deviation from equilibrium, while the voltage across the capacitor $C_2$ is

U_{20} = \frac{q_{20}}{C_2} = \frac{C_1}{C_1 + C_2}U_0. \qquad (16)

Note that the zero coil current corresponds to the initial moment when the key $K$ is just shorted.

After a quarter of a period has passed, the current in the coil becomes maximum and the system passes the equilibrium position, whereas the voltage across the capacitor $C_2$ drops to the value

U_2 = \frac{q_2}{C_2} = \frac{C_1}{C_1 + C_2 + C_3}U_0, \qquad (17)

that is, it falls by $U_{20} - U_2$ . After another quarter of the period, the voltage across the capacitor will further drop by the same amount, which is, at the same time, equal to $U_2 - U_{\min}$ , so the minimum voltage is ultimately obtained as

U_{\min} = 2U_2 - U_{20} = \frac{C_1(C_1 + C_2 - C_3)}{(C_1 + C_2)(C_1 + C_2 + C_3)}U_0. \qquad (18)

Content	Points
Formula (1): $q_{30} = 0$	0.2
Formula (2): $q_{10} = q_{20}$	0.2
Formula (3): $\dfrac{q_{10}}{C_1} + \dfrac{q_{20}}{C_2} = U_0$	0.2
Formula (4): $q_{10} = q_{20} = \dfrac{C_1C_2}{C_1 + C_2}U_0$	0.2
Formula (5): $W_0 = \dfrac{q_{10}^2}{2C_1} + \dfrac{q_{20}^2}{2C_2} = \dfrac{C_1C_2U_0^2}{2(C_1 + C_2)}$	0.2
Formula (6): $q_1 = q_2 + q_3$	0.2
Formula (7): $\dfrac{q_2}{C_2} = \dfrac{q_3}{C_3}$	0.2
Formula (8): $\dfrac{q_1}{C_1} + \dfrac{q_2}{C_2} = U_0$	0.2
Formula (10): $q_2 = \dfrac{C_1C_2}{C_1 + C_2 + C_3}U_0$	0.2
Formula (12): $W = \dfrac{C_1(C_2 + C_3)U_0^2}{2(C_1 + C_2 + C_3)} + \dfrac{LI_{\max}^2}{2}$	0.2
Formula (13): $A = (q_1 - q_{10})U_0$	0.2
Formula (14): $W_0 + A = W$	0.2
Formula (15): $I_{\max} = \sqrt{\dfrac{C_3}{(C_1 + C_2)(C_1 + C_2 + C_3)L}}C_1U_0$	0.2
Formula (18): $U_{\min} = 2U_2 - U_{20}$	0.2
Formula (18): $U_{\min} = \dfrac{C_1(C_1 + C_2 - C_3)}{(C_1 + C_2)(C_1 + C_2 + C_3)}U_0$	0.2
Total	3.0

Mathematical hints for the theoretical problems

The following formulas may be useful:

\int x^ndx = \frac{x^{n+1}}{n+1}, \text{ where } n \text{ is an integer;}

(1 + x)^\gamma \approx 1 + \gamma x + \frac{\gamma(\gamma - 1)}{2}x^2, \textit{ for } x \ll 1 \textit{ and any } \gamma.

Оригинал в Архива: IZhO-2021-Theory_eng.pdf · официални решения: IZhO-2021-Theory_eng_sol.pdf