IZhO 2019, Теория — Задача 1

• 1A $t = \frac{l_0^2T^2 - m^2v^4}{2RvT^2}$

• 1B 28 °C

  Official answer: $T_x = 301K = 28^{\circ}C$; full marks for 27.5–28.5 °C.

• 1C A resistor of 2019 Ohms in series with a coil of inductance 1.00·10⁻² Hn; maximum power 13.9 mW.

  Official: $L = \frac{1}{\omega^2C} = 1.00\cdot 10^{-2}$ Hn, $P_{max} = \frac{U^2}{8r} = 13.9$ mW.

Problem 1A (4.0 points)

Since the thread is inextensible and under stress, then the speed of the puck is always perpendicular to the thread. Therefore, the tension force of the thread does not perform any work on the puck and its speed remains constant by modulus

The puck moves along its trajectory with the curvature radius equal to the length $l$ of the unwound thread, therefore, the condition for the thread to be torn up is found from Newton's second law as

The length of the thread changes as a result of winding on the cylinder according to

where

and the angular velocity of the thread rotation is obtained as follows

It follows from equations (3)-(5) that

and its integration entails

Substituting formula (1) into (7), the time moment sought is finally found as

Problem 1B (3.0 points)

Possible solution. The power of the heat transfer from the body to the air is proportional to the difference between the body $T$ and the air $T_x$ temperatures with the factor $\alpha$, i.e.

as a result, the body with the heat capacity $C$ cools down by the temperature $dT$ over time period $dt$, which obeys the heat balance equation

Equations (1) and (2) with the initial condition $T = T_0$ have a solution

where $\beta = \alpha / C$ is a constant.

Let the body be cooled from the temperature $T_0$ to the temperature $T_1$ for a certain time interval, then it follows from (3) that

where $\gamma$ is a constant.

Over the following same time interval, this difference will also change in $\gamma$ times

Equations (4) and (5) result in the relation

which has the following solution

It is obtained from the graph provided: the initial temperature $T_0 = 373\mathrm{K}$, in 10 minutes the temperature is equal to $T_1 = 337K$, and in 20 minutes it reaches the value of $T_2 = 319K$. Substituting these data into equation (7), the air temperature is finally calculated as

Problem 1C (3.0 points)

Let $R$ be the active component of the load (the real part of the impedance), and $X$ be the reactive component of the entire circuit (the imaginary part of the total impedance). Then the current amplitude is found as

The average thermal power in the load reads as

It is seen that the maximum power is achieved at $X = 0$, i.e. there should be no phase shift in the circuit. The remaining expression has a maximum at $R = r$.

The phase shift would be zero if a coil was connected in series with the capacitor such that $\frac{1}{\omega C} = \omega L$, and, thus, $L = \frac{1}{\omega^2C} = 1.00 \cdot 10^{-2}\ Hn$.

It turns out that the simplest load must consist of the resistor with the resistance of 2019 Ohms and the coil with the inductance of $1.00 \cdot 10^{-2}\ Hn$.

The maximum power is obtained as