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Условие Torsion: Construction of the Potential Curve Отговори Решение

IZhO 2026, Експеримент — Задача 1. Torsion: Construction of the Potential Curve

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XXII International Zhautykov Olympiad/Experimental competition — EXPERIMENTAL COMPETITION · 12 януари 2026 г.

Внимание: Бележка към темата

The question refers to an approximate formula for $T(N)$ after Part 1.2, but no formula is printed there before Part 1.3. The writing sheets referenced by the question are not present in this retained question/key packet.

Условие

Please read the instructions first:

The Experimental competition consists of one problem. This part of the competition lasts 4 hours.
Please only use the pen that is provided to you.
You can use your own non-programmable 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 papers. 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 bordered area.
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 (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:

Used Writing sheets in order.
The sheets you do not wish to be evaluated.
Unused sheets.
The printed problems.

Place the papers inside the envelope and leave everything on your desk. You are not allowed to take any paper or equipment out of the room.

Torsion: Construction of the Potential Curve

The behavior of mechanical systems with one degree of freedom is completely determined by the dependence of the system's potential energy on the coordinate, the graph of which is called the potential curve. In many cases, a theoretical calculation of this dependence can be complicated and sometimes even impossible. In such situations, the potential curve can be obtained on the basis of experimental data.
In this work, you are required to determine the dependence of the potential energy on the coordinate for the free rotation of a metal rod suspended by two vertical threads. During all experiments, keep the thread length $l$ and the distance between them $h$ constant.

Four side-view diagrams of a horizontal rod suspended by two threads from a fixed support. Diagram 1: vertical parallel threads with length l and spacing h labelled, arrow showing rotation of the rod. Diagram 2: threads crossed into contact, labelled N = 0. Diagram 3: threads twisted together, labelled N, t = 0. Diagram 4: crossed threads with rod at lower position, labelled N = 0, t = T.

To carry out the measurements, use the following procedure. Starting from the initial position, when the threads are vertical (1), rotate the rod by half a turn (2), so that the threads come into contact. Then continue rotating the rod so that the threads are twisted, making the required number of full turns $N$ (3). After that, rotate the rod by an additional quarter turn and release it. Start the stopwatch only when, during unwinding, the rod again reaches position (3). After that, measure the time for the required number of rotation turns until position (4). In the calculations, assume that the angular velocity of the rod in the initial position (3) is zero, despite the additional quarter turn.
In this problem, two interrelated coordinates are used: $N$ — the number of turns during the twisting of the threads, counted from the lower position of the rod (2) upward; $k$ — the number of turns made by the rod during the unwinding of the threads, counted from position (3) at $N = 30$ downward. The unit of measurement for the coordinates is "one turn." Accordingly, the unit of velocity $V$ is "number of turns per second," with dimension $[V] = \mathrm{s}^{-1}$ . As a measure of energy, the square of the velocity $E = V^2$ is used, which is proportional to the rotational kinetic energy of the rod. Let us call the unit of energy in this case $[E] = \mathrm{s}^{-2}$ — Ku (Kazakhstani unit). The potential energy of the rod is taken to be zero at position (3) for $N = 30$ , unless another "zero" position is specified.

Vertical ruler-like scale: on the left the axis N (arrow upward) with ticks 0, 5, 10, 15, 20, 25, 30; on the right the axis k (arrow downward) with ticks 0, 5, 10, 15; the mark N = 30 coincides with k = 0.

Part 1. Theoretical Introduction

The mass of the rod is $m_1 = 50.0$ g, and the mass of one nut is $m_2 = 6.4$ g.

During the measurements, you need to measure the times $t(k)$ over which the rod makes $k$ turns. Let this dependence be approximately described by the formula

t(k) = Ak^{\alpha},

where $A, \alpha$ are constants.

Let the potential energy $U$ be taken as zero at position (2), and let its dependence on the coordinate $N$ be approximately described by the formula:

U(N) = BN^{\beta}.

Denote by $T$ the time required for the rod to unwind from the initial position (3) to the lower position (4). Under these conditions, the approximate dependence $T(N)$ is given by the formula

Part 2. Study of the Law of Motion

Set the rod in the initial position (3) with $N = 30$ . Release the rod and use a stopwatch with lap memory to record the times $t(k)$ during which the rod makes $k$ turns from the initial position. Perform the measurements in the range $k = 0$ to 30, taking data every three turns.

To calculate the rod's velocity from the experimental data, use the symmetric formula

V(k) = \frac{6}{t(k+3)-t(k-3)}.

Part 3. Step-by-Step Probing

In this part, estimation of uncertainties is not required. To determine the parameters of the dependences, use the graphical method.
The smaller the interval, the more accurately the approximate (fitting) formulas describe the motion. In this part of the work, you are required to study the motion of the rod over small intervals of variation of the coordinate $N$ , the boundaries of which are specified in the Writing Sheets. The larger value corresponds to the initial position. For each interval, you must measure the law of motion, namely, the times during which the rod makes $k = 10$ turns from the initial position. Within each given interval of the coordinate $N$ , the coordinate $k$ varies from 1 to 10. Perform the measurements with a step $\Delta k = 1$ . Use a stopwatch with lap memory and record the time after each turn of the rod.
For each interval, assume that the obtained dependence $t(k)$ can be approximately described by the function $t(k) = Ak^{\alpha}$ with its own values of the parameters $A$ and $\alpha$ , which you must determine. For each of the specified intervals, perform the following tasks.

Summarize the obtained results. To this end, assume that the rod unwinds from the initial position $N = 30$ .

Part 4. Unwinding Time

In this part, you are required to investigate the dependence of the total unwinding time $T(N)$ of the rod on the initial value $N$ until the lower position (4).

Assume that the unwinding of the rod starts from the position $N = 30$ .

Choose the value of the coefficient $B$ in the formula $U = BN^{\beta}$ such that the value of the kinetic energy at $k = 30$ coincides with the value of this energy calculated in Section 3.6.

1.1 Express the energy unit Ku in joules and calculate its numerical value.

1.2 Derive a formula for calculating the velocity $V(k)$ and kinetic energy $E(k)$ in Ku units as a function of the coordinate $k$ , which may be formally treated as continuous. Express them in terms of the parameters $A$ , $\alpha$ , and $k$ .

1.3 Express the exponent $\gamma$ in terms of the exponent $\beta$ .

2.1 Enter the measured values $t(k)$ into Table 1.

2.2 Plot the graph of the law of motion $k(t)$ .

2.3 Using the experimental data, calculate the values of the kinetic energy $E(k)$ for all measured values of the coordinate $k$ . Enter the results into Table 1 in the column "E(k) (exp.)."

2.4 Plot the resulting dependence $E(k)$ . Label it as No. 1.

2.5 Write down the formula for the potential energy of the rod $U(k)$ , expressing it in terms of $E(k)$ .

2.6 Plot the dependence of $t(k)$ on the number of turns $k$ using a logarithmic scale.

2.7 Calculate the values of the parameters $A$ and $\alpha$ , and estimate their uncertainties.

2.8 Using the obtained values of the parameters $A$ and $\alpha$ , calculate the kinetic energy values $E(k)$ in accordance with Section 1.2. Enter the results into Table 1 in the column "E(k) (theor.)."

2.9 Plot the calculated dependence of the rod's kinetic energy $E(k)$ on the same graph as in Section 2.4. Label it as No. 2.

3.1 Measure the values of the times $t(k)$ during which the rod makes $k$ turns from the initial position.

3.2 Plot the dependence $t(k)$ on the prepared sections of the Writing Sheets using a logarithmic scale.

3.3 Draw a smoothing straight line through the last six points (for values of $k$ from 5 to 10).

3. Using the constructed linear graph, determine the values of the parameters $A$ and $\alpha$ , indicating the formulas used for their calculation.

3.5 Calculate the change in the kinetic energy of the rod $E_{5-10} = E(10) - E(5)$ as the coordinate $k$ changes from 5 to 10. Provide the formula used to calculate this quantity.

3.6 Using the data obtained in this part, calculate the values of the rod's kinetic energy $E(k)$ after $k = 5, 10, 15, 20, 25,$ and $30$ turns. Plot, on the graph from Section 2.4, the dependence obtained in this part. Label it as No. 3.

4.1 Measure the dependence of the total unwinding time $T(N)$ of the rod on the number of turns $N$ for $N = 5, 10, 15, 20, 25, 30..$

4.2 Estimate the uncertainty in the measurement of the unwinding time for $N = 10$ . To do this, perform at least five measurements of this time.

4.3 Plot the obtained dependence $T(N)$ using a logarithmic scale.

4.4 Assuming that this dependence is described by the formula $T(N) = GN^{\gamma}$ , determine the exponent $\gamma$ in this formula.

4.5 Using the obtained value of the exponent $\gamma$ , calculate the value of the exponent $\beta$ in the formula for the dependence of the potential energy on the coordinate, $U = BN^{\beta}$ .

4.6 Calculate the values of the rod's kinetic energy $E(k)$ after $k = 5, 10, 15, 20, 25, 30$ turns. Provide the formulas that you used to calculate $E(k)$ .

4.7 Plot, on the graph from Section 2.4, the dependence obtained in this part. Label it as No. 4.

Отговори

Решение

SOLUTION TO THE EXPERIMENTAL COMPETITION

Torsion: Construction of the Potential Curve

Part 1. Theoretical Introduction

1.1 The kinetic energy of a rotating rod is given by the formula

E = \frac{I\omega^2}{2}.

(1)

Here $I = \frac{m_1 a^2}{12} + 2m_2\frac{h^2}{4}$ is the moment of inertia of the rod with respect to the rotation axis (where $a$ is the length of the rod and $h$ is the distance between the threads), and $\omega = 2\pi V$ is the angular velocity of the rod ( $\omega$ is the rotation frequency).

Therefore,

E = \pi^2\left(\frac{m_1 a^2}{6} + m_2 h^2\right)V^2

(2)

At $V = 1\ \mathrm{s^{-1}}$ , the energy of the rod is equal to the unit of energy sought. Here and in what follows, it is assumed that the energy associated with the vertical velocity of the rod can be neglected in comparison with the energy of rotational motion. According to the measurement results, $a = 25\ \mathrm{cm}$ , $h = 15\ \mathrm{cm}$ , and the mass values are given in the problem statement. Therefore,

1\ Ku \approx 6.35\cdot 10^{-3}\ \mathrm{J}.

(3)

1.2 Let us calculate the derivative of the proposed dependence of time on the coordinate, $t(k) = Ak^{\alpha}$ , with respect to $k$ , taking into account that the coordinate is the number of turns $k$ (not necessarily an integer):

V(k) = \frac{dk}{dt} = \frac{k^{1-\alpha}}{\alpha A},

(4)

E(k) = V^2 = \left(\frac{k^{1-\alpha}}{\alpha A}\right)^2.

(5)

1.3 Let us express the angular velocity of the rod from the law of energy conservation:

V^2 = U(N) - U(k).

(6)

The unwinding time can be calculated using the following formula:

T = \int_0^N \frac{dk}{V(k)}.

(7)

Introducing a change of the integration variable,

\xi = \frac{N-k}{N}

(8)

and using the dependence $U(N) = BN^{\beta}$ , we obtain

T(N) = N^{1-\frac{\beta}{2}}\sqrt{\frac{1}{B}}\int_0^1 \frac{d\xi}{\sqrt{1-\xi^{\beta}}}.

(9)

The integral in this expression does not depend on the value of $N$ ; it depends only on the exponent $\beta$ . Therefore, the dependence of the unwinding time on the number of turns is given by the formula $T(N) = GN^{\gamma}$ , in which the exponent is

\gamma = 1 - \frac{\beta}{2}.

(10)

Part 2. Study of the Law of Motion

2.1-2.2 The results of the measurements and the required calculations are presented in Table 1.

$k$	$t$ , s	$V$ , s $^{-1}$	$E$ , Ku (exp.)	$\ln k$	$\ln t$	$E$ , Ku (theor.)
0	0	0	0
3	9,55	0,391	0,153	1,099	2,257	0,213
6	15,35	0,568	0,323	1,792	2,731	0,330
9	20,11	0,662	0,438	2,197	3,001	0,427
12	24,42	0,725	0,525	2,485	3,195	0,512
15	28,39	0,774	0,599	2,708	3,346	0,590
18	32,17	0,803	0,645	2,890	3,471	0,662
21	35,86	0,823	0,677	3,045	3,580	0,729
24	39,46	0,838	0,702	3,178	3,675	0,794
27	43,02	0,852	0,726	3,296	3,762	0,855
30	46,5			3,401	3,839	0,914

The figure below shows the graph of the coordinate as a function of time.

Graph k(t): smoothly curving line with experimental points from k = 0 to k = 30 for t from 0 to ~46.5 s; axes are t, s (0–60) and k (0–33).

2.3 able 1 shows the values of the rod's velocities $V$ , calculated using the formula given in the problem statement. The kinetic energy values were calculated using the formula $E = V^2$ .

2.4 The graph of the resulting dependence is shown in the figure (No. 1).

Graph of E(k), Ku versus k (0–33) with four series of points and smoothing curves labelled №1, №2, №3, №4; E rises from 0 to about 0,86 Ku.

2.5 he potential energy at the specified zero level is

U(k) = -E(k).

(11)

2.6 The graph of the dependence of the motion time on the number of turns is shown in the figure below, plotted on a logarithmic scale.

Graph of ln t versus ln k (ln k from 0,0 to 3,5; ln t from 2,0 to 4,0) with the experimental points and the straight line y = 0,684x + 1,500.

2. 7 The coefficients of this dependence, $\ln t = \alpha \ln k + \ln A$ , calculated using the least squares method, are

\alpha = 0.684 \pm 0.006,

(12)

\ln A = 1.50 \pm 0.02.

(13)

The calculation of the parameter $A$ and its error yields

A = \exp(\ln A) = 4.48\ Ku,

(14)

\Delta A = A\Delta(\ln A) = \pm 0.07\ Ku.

(15)

2.8-2.9 To calculate the kinetic energy, use formula (5) with the obtained numerical values of the parameters. The results of the calculations are presented in Table 1 and on the graph from Section 2.4, labeled as No. 2.

Part 3. Step-by-Step Probing

3.1-3.5 The results of the measurements, calculations, graphs, and the final result (calculated value of the energy change $E_{5-10}$ ) for each interval are presented below.

The values of the parameter $\alpha$ are calculated as the slope of the graph:

\alpha = \frac{\Delta(\ln t)}{\Delta(\ln k)}.

(16)

It is convenient to take the values of the extreme points on the graph (the points where the line intersects the graph boundaries), denoted as $z_{\min}$ and $z_{\max}$ , while $\Delta(\ln k) = 2.5$ .

The value of the parameter $\ln A$ corresponds to the coordinate of the intersection of the line with the ordinate axis.

To calculate the change in kinetic energy, the following formula is used:

E_{5-10} = \left(\frac{10^{1-\alpha}}{\alpha A}\right)^2 - \left(\frac{5^{1-\alpha}}{\alpha A}\right)^2 = \left(\frac{5^{1-\alpha}}{\alpha A}\right)^2\left(4^{1-\alpha} - 1\right).

(17)

Interval 35-25

$k$	$t$ , s	$\ln k$	$\ln t$
0	0,00
1	4,14	0,000	1,421
2	6,96	0,693	1,940
3	9,19	1,099	2,218
4	11,29	1,386	2,424
5	13,17	1,609	2,578
6	14,94	1,792	2,704
7	16,60	1,946	2,809
8	18,09	2,079	2,895
9	19,63	2,197	2,977
10	21,04	2,303	3,046

$\alpha = 0{,}672$ ; $z_{\max} = 3{,}18$ ; $z_{\min} = 1{,}50$ ; $\ln A = 1{,}500$ ; $A = 4{,}482$ .

E_{5-10} = 0{,}182\ Ku

Interval 35-25: graph of ln t versus ln k (0,0–2,5; 1,0–3,6) with points and the fit line y = 0,6748x + 1,4938.

Interval 30-20

$k$	$t$ , s	$\ln k$	$\ln t$
0	0
1	4,43	0,000	1,488
2	7,43	0,693	2,006
3	9,92	1,099	2,295
4	11,96	1,386	2,482
5	13,94	1,609	2,635
6	15,71	1,792	2,754
7	17,47	1,946	2,860
8	19,06	2,079	2,948
9	20,61	2,197	3,026
10	22,09	2,303	3,095

$\alpha = 0{,}664$ ; $z_{\max} = 3{,}23$ ; $z_{\min} = 1{,}57$ ; $\ln A = 1{,}570$ ; $A = 4{,}807$ .

E_{5-10} = 0{,}172\ Ku

Interval 25-15

$k$	$t$ , s	$\ln k$	$\ln t$
0	0,00
1	4,73	0,000	1,554
2	7,85	0,693	2,061
3	10,44	1,099	2,346
4	12,69	1,386	2,541
5	14,73	1,609	2,690
6	16,59	1,792	2,809
7	18,41	1,946	2,913
8	20,20	2,079	3,006
9	21,87	2,197	3,085
10	23,48	2,303	3,156

$\alpha = 0{,}680$ ; $z_{\max} = 2{,}23$ ; $z_{\min} = 0{,}57$ ; $\ln A = 1{,}600$ ; $A = 4{,}953$ .

E_{5-10} = 0{,}138\ Ku

Interval 25-15: graph of ln t versus ln k (0,0–2,5; 1,0–3,8) with points and the fit line y = 0,6751x + 1,6012.

Interval 20-10

$k$	$t$ , s	$\ln k$	$\ln t$
0	0
1	5,01	0,000	1,611
2	8,38	0,693	2,126
3	11,19	1,099	2,415
4	13,67	1,386	2,615
5	15,92	1,609	2,768
6	17,98	1,792	2,889
7	19,99	1,946	2,995
8	21,86	2,079	3,085
9	23,61	2,197	3,162
10	25,39	2,303	3,234

$\alpha = 0{,}672$ ; $z_{\max} = 3{,}30$ ; $z_{\min} = 1{,}60$ ; $\ln A = 1{,}680$ ; $A = 5{,}366$ .

E_{5-10} = 0{,}127\ Ku

Interval 15-5

$k$	$t$ , s	$\ln k$	$\ln t$
0	0
1	5,55	0,000	1,714
2	9,28	0,693	2,228
3	12,33	1,099	2,512
4	15,08	1,386	2,713
5	17,53	1,609	2,864
6	19,86	1,792	2,989
7	22,04	1,946	3,093
8	24,13	2,079	3,183
9	26,07	2,197	3,261
10	28,05	2,303	3,334

$\alpha = 0{,}676$ ; $z_{\max} = 3{,}47$ ; $z_{\min} = 1{,}78$ ; $\ln A = 1{,}780$ ; $A = 5{,}930$ .

E_{5-10} = 0{,}100\ Ku

Interval 15-5: graph of ln t versus ln k (0,0–2,5; 1,0–3,8) with points and the fit line y = 0,6765x + 1,7759.

Interval 10-0

$k$	$t$ , s	$\ln k$	$\ln t$
0	0
1	6,48	0,000	1,869
2	10,74	0,693	2,374
3	14,16	1,099	2,650
4	17,16	1,386	2,843
5	19,94	1,609	2,993
6	22,44	1,792	3,111
7	24,84	1,946	3,212
8	27,08	2,079	3,299
9	29,28	2,197	3,377
10	31,23	2,303	3,441

$\alpha = 0{,}648$ ; $z_{\max} = 2{,}57$ ; $z_{\min} = 0{,}95$ ; $\ln A = 1{,}950$ ; $A = 7{,}029$ .

E_{5-10} = 0{,}094\ Ku

3.6 The final table is as follows:

Upper boundary $N$	Lower boundary $N$	$E_{5-10}$ , $Ku$	$N$	$k$	$E(k)$
35	25	0,182		0	0
30	20	0,172	30	5	0,182
25	15	0,138	25	10	0,354
20	10	0,127	20	15	0,492
15	5	0,100	15	20	0,619
10	0	0,094	10	25	0,719
			5	30	0,814

The values of $E(k)$ were obtained by stepwise summation:

E_0 = 0

E(k+5) = E(k) + E_{5-10}

(18)

The graph of this dependence, labeled No. 3, is shown on the graph from Section 2.4.

Part 4. Unwinding Time

4.1 The table of measurement results is shown below:

$N$	$T$ , s	$\ln N$	$\ln\ T$	$N$	$k$	$E(k)$
30	47,63	3,40	3,86		0	0,000
25	45,30	3,22	3,81	30	5	0,149
20	42,62	3,00	3,75	25	10	0,294
15	37,25	2,71	3,62	20	15	0,436
10	30,81	2,30	3,43	15	20	0,572
5	21,65	1,61	3,07	10	25	0,702
				5	30	0,814

4.2. To estimate the error in the measurement of the unwinding time, several measurements must be performed. The random error is calculated strictly using the formula

\Delta T = t_{n,p}\sqrt{\frac{\sum_{i=1}^{n}(x_i - \langle x\rangle)^2}{n(n-1)}}

(19)

and amounts to a few hundredths of a second.

4.3 The graph of the dependence of the unwinding time on the initial number of turns is shown in the figure.

Graph of ln T versus ln N (ln N from 1,5 to 3,5; ln T from 3,0 to 4,0) with six measured points and the fit line y = 0,4473x + 2,3811.

4.4 The value of the exponent can be determined either from the graph or by the least squares method. Its value is

\gamma = 0.45.

(20)

4.5 The exponent in the formula for the potential energy is

\beta = 2(1 - \gamma) = 1.10.

(21)

4.6 To calculate the kinetic energy, perform the following steps:

a) Using the formula $U' = N^{\beta}$ , calculate the values of the potential energy;

b) Using these values, calculate the values of the kinetic energy (up to a constant factor):

E'(k) = U'(30) - U'(N-k).

(22)

в) Normalize the energy:

E(k) = \frac{E(30)}{E'(30)}E'(k),

(23)

where $E(30)$ is the value of the kinetic energy found in Part 3 of the assignment (from our measurements, $E(30) = 0.814$ ).

The results of the calculations are presented in Table 4.1. The graph of this dependence, labeled No. 4, is shown on the graph from Section 2.4.

Marking scheme

Correct calculations using incorrect formulas will not be graded!
Incorrect rounding will result in a penalty (-0.2 points).

Part	Content	Points	Total for part
Part 1 Theoretical Introduction
1.1	Formula (2): $E = \pi^2\left(\frac{m_1 a^2}{6} + m_2 h^2\right)V^2$	0,2	0,5
1.1	Measured: $a = (25 \pm 1)\,\mathrm{sm}$ and $h = (15 \pm 1)\,\mathrm{sm}$	0,2
1.1	Numerical value in formula (3): $1\ Ku \approx 6.35\cdot 10^{-3}\ \mathrm{J}$	0,1
1.2	Formula (4): $V(k) = \frac{k^{1-\alpha}}{\alpha A}$	0,3	0,5
1.2	Formula (5): $E(k) = \left(\frac{k^{1-\alpha}}{\alpha A}\right)^2$	0,2
1.3	Integral (7): $T = \int_0^N \frac{dk}{V(k)}$	0,2	0,6
1.3	Formula (9): $T(N) = N^{1-\frac{\beta}{2}}\sqrt{\frac{1}{B}}\int_0^1 \frac{d\xi}{\sqrt{1-\xi^{\beta}}}$	0,3
1.3	Exponent in formula (10): $\gamma = 1 - \frac{\beta}{2}$	0,1
Part 2. Study of the Law of Motion
2.1	Measurements are made 0,1x10; within the interval 10% 0,1x10 (within the interval 20% 0,05x10)	1	1
Parts 2.2 – 2.4 and 2.6 – 2.9 are graded only if part 2.1 is graded
2.2	Plotting graph $k(t)$ : the data points are plotted according to the table.	0,2	0,3
2.2	smoothing curve is drawn	0,1
2.3	Calculations of the kinetic energy are performed	0,4	0,4
2.4	Plotting the graph $E(k)$ : the data points are plotted according to the table.	0,2	0,3
2.4	smoothing curve is drawn	0,1
2.5	Formula (11) for the potential energy: $U(k) = -E(k)$	0,2	0,2
2.6	Plotting the graph $\ln t$ versus $\ln k$ : the logarithms of the times are calculated	0,1	0,4
2.6	the data points are plotted according to the table	0,2
2.6	best-fit straight line is drawn	0,1
2.7	Determination of the parameters of the dependence. LSM is used	0,2	0,9
2.7	numerical value within the range $\alpha = 0{,}68 \pm 0{,}05$ (within the range $\alpha = 0{,}68 \pm 0{,}10$ — 0,1)	0,2
2.7	error $\Delta\alpha$	0,1
2.7	numerical value within the range $A = 4{,}5 \pm 0{,}3$ (within the range $A = 4{,}5 \pm 0{,}6$ — 0,1)	0,2
2.7	Error of $A$ (formula → numerical value)	0,2
2.8	Calculation of the theoretical energy values of $E(k)$		0,5
2.9	Plotting the graph $E(k)$ (graded only if part.2.8 is graded): the data points are plotted according to the table	0,2	0,3
2.9	smoothing curve is drawn	0,1
Part 3. Step-by-Step Probing — At all intervals, parts 3.2–3.5 are graded only if the measurement results of the corresponding part 3.1 is graded.
Interval 35-25
3.1	Measurements are made. Within the range 10% — 0,3 (within the range 20% — 0,1)		0,3
3.2	Plotting the graph: the logarithms $\ln t$ are calculated	0,1	0,2
3.2	the data points are plotted according to the table	0,1
3.3	best-fit straight line is drawn		0,1
3.4	Calculation of the parameters. Formula for $\alpha$ : $\alpha = \frac{\Delta(\ln t)}{\Delta(\ln k)}$	0,1	0,8
3.4	Formula for $A$ : $A = \exp(\ln A)$	0,1
3.4	numerical value within the range $\alpha = 0{,}67 \pm 0{,}06$ (within the range $\alpha = 0{,}67 \pm 0{,}10$ — 0,1)	0,3
3.4	numerical value within the range $A = 4{,}5 \pm 0{,}03$ (within the range $A = 4{,}5 \pm 0{,}06$ — 0,1)	0,3
3.5	Calculation of energy. Formula (17): $E_{5-10} = \left(\frac{10^{1-\alpha}}{\alpha A}\right)^2 - \left(\frac{5^{1-\alpha}}{\alpha A}\right)^2 = \left(\frac{5^{1-\alpha}}{\alpha A}\right)^2\left(4^{1-\alpha} - 1\right)$	0,2	0,5
3.5	numerical value within the range $E_{5-10} = 0{,}18 \pm 0{,}06$ (within the range $E_{5-10} = 0{,}18 \pm 0{,}10$ — 0,1)	0,3
Interval 30-20
3.1	Measurements are made. within the range 10% — 0,3 (within the range 20% — 0,1)		0,3
3.2	Plotting the graph: the logarithms $\ln t$ are calculated	0,1	0,2
3.2	the data points are plotted according to the table	0,1
3.3	best-fit straight line is drawn		0,1
3.4	Calculation of the parameters. numerical value within the range $\alpha = 0{,}66 \pm 0{,}06$ (within the range $\alpha = 0{,}66 \pm 0{,}10$ — 0,1)	0,3	0,6
3.4	numerical value within the range $A = 4{,}8 \pm 0{,}03$ (within the range $A = 4{,}8 \pm 0{,}06$ — 0,1)	0,3
3.5	numerical value within the range $E_{5-10} = 0{,}17 \pm 0{,}06$ (within the range $E_{5-10} = 0{,}17 \pm 0{,}10$ — 0,1)		0,3
Interval 25-15
3.1	Measurements are made. Within the range 10% — 0,3 (within the range 20% — 0,1)		0,3
3.2	Plotting the graph: the logarithms $\ln t$ are calculated	0,1	0,2
3.2	the data points are plotted according to the table	0,1
3.3	best-fit straight line is drawn		0,1
3.4	Calculation of the parameters. numerical value within the range $\alpha = 0{,}68 \pm 0{,}06$ (within the range $\alpha = 0{,}68 \pm 0{,}10$ — 0,1)	0,3	0,6
3.4	numerical value within the range $A = 4{,}95 \pm 0{,}03$ (within the range $A = 4{,}95 \pm 0{,}06$ — 0,1)	0,3
3.5	numerical value within the range $E_{5-10} = 0{,}14 \pm 0{,}06$ (within the range $E_{5-10} = 0{,}14 \pm 0{,}10$ — 0,1)		0,3
Interval 20-10
3.1	Measurements are made. Within the range 10% — 0,3 (within the range 20% — 0,1)		0,3
3.2	Plotting the graph: the logarithms $\ln t$ are calculated	0,1	0,2
3.2	the data points are plotted according to the table	0,1
3.3	best-fit straight line is drawn		0,1
3.4	Calculation of the parameters. numerical value within the range $\alpha = 0{,}67 \pm 0{,}06$ (within the range $\alpha = 0{,}67 \pm 0{,}10$ — 0,1)	0,3	0,6
3.4	numerical value within the range $A = 5{,}36 \pm 0{,}03$ (within the range $A = 5{,}36 \pm 0{,}06$ — 0,1)	0,3
3.5	numerical value within the range $E_{5-10} = 0{,}13 \pm 0{,}06$ (within the range $E_{5-10} = 0{,}13 \pm 0{,}10$ — 0,1)		0,3
Interval 15-5
3.1	Measurements are made. Within the range 10% — 0,3 (within the range 20% — 0,1)		0,3
3.2	Plotting the graph: the logarithms $\ln t$ are calculated	0,1	0,2
3.2	the data points are plotted according to the table	0,1
3.3	best-fit straight line is drawn		0,1
3.4	Calculation of the parameters. numerical value within the range $\alpha = 0{,}67 \pm 0{,}06$ (within the range $\alpha = 0{,}67 \pm 0{,}10$ — 0,1)	0,3	0,6
3.4	numerical value within the range $A = 5{,}93 \pm 0{,}03$ (within the range $A = 5{,}93 \pm 0{,}06$ — 0,1)	0,3
3.5	numerical value within the range $E_{5-10} = 0{,}10 \pm 0{,}06$ (within the range $E_{5-10} = 0{,}10 \pm 0{,}10$ — 0,1)		0,3
Interval 10-0
3.1	Measurements are made. Within the range 10% — 0,3 (within the range 20% — 0,1)		0,3
3.2	Plotting the graph: the logarithms $\ln t$ are calculated	0,1	0,2
3.2	the data points are plotted according to the table	0,1
3.3	best-fit straight line is drawn		0,1
3.4	Calculation of the parameters. numerical value within the range $\alpha = 0{,}65 \pm 0{,}06$ (within the range $\alpha = 0{,}65 \pm 0{,}10$ — 0,1)	0,3	0,6
3.4	numerical value within the range $A = 7{,}03 \pm 0{,}03$ (within the range $A = 7{,}03 \pm 0{,}06$ — 0,1)	0,3
3.5	numerical value within the range $E_{5-10} = 0{,}09 \pm 0{,}06$ (within the range $E_{5-10} = 0{,}09 \pm 0{,}10$ — 0,1)		0,3
3.6	Calculation of the energy dependence on the coordinate. Formulas (18): $E_0 = 0$ ; $E(k+5) = E(k) + E_{5-10}$	0,2	0,9
3.6	The energies are calculated.	0,4
3.6	Construction of Graph No. 3.	0,3
Part 4. Unwinding Time
4.1	Measurements are made 0,2x6; within the range 10% 0,2x6 (within the range 20% 0,1x6)		1,2
Parts 4.3–4.5 and 4.7 are graded only if the measurement results in part 4.1 are graded
4.2	Estimation of the time measurement error. 5–7 measurements are taken (acceptable error — 10%) (3-4 measurements – 0,1)	0,3	0,6
4.2	average is calculated	0,1
4.2	error is estimated using any correct formula	0,2
4.3	Plotting the graph $T(N)$ : logarithms $\ln T$ are calculated	0,1	0,4
4.3	data points are plotted according to the table	0,2
4.3	best-fit straight line is drawn	0,1
4.4	exponent is obtained $\gamma = 0{,}45 \pm 0{,}03$ (within the range $\gamma = 0{,}45 \pm 0{,}06$ - 0,1)		0,3
4.5	The value is calculated $\beta = 2(1-\gamma)$		0,2
4.6	Energy calculations; calculation procedure		0,8
4.6	calculation of the potential energy	0,1
4.6	calculation of the kinetic energy (22): $E'(k) = U'(30) - U'(N-k)$	0,1
4.6	normalization according to Eq. (23): $E(k) = \frac{E(30)}{E'(30)}E'(k)$	0,1
4.6	energy calculations are made	0,5
4,7	Plotting graph №4	0,3	0,3
TOTAL			20

Оригинал в Архива: IZhO-2026-Exp_eng.pdf · официални решения: IZhO-2026-Exp_eng_sol.pdf