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IPhO 2014, Теория — Задача 1. Problem 1

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Theoretical competition. Tuesday, 15 July 2014 · 9 т.

Условие

This problem consists of three independent parts.

Part A A small puck of mass $m$ is carefully placed onto the inner surface of the thin hollow thin cylinder of mass $M$ and of radius $R$ . Initially, the cylinder rests on the horizontal plane and the puck is located at the height $R$ above the plane as shown in the figure on the left. Find the interaction force $F$ between the puck and the cylinder at the moment when the puck passes the lowest point of its trajectory. Assume that the friction between the puck and the inner surface of the cylinder is absent, and the cylinder moves on the plane without slipping. The free fall acceleration is $g$ . [3 т.]

A thin hollow cylinder labelled M, R resting on a hatched horizontal plane; a small square puck m sits on the inner surface at the left end of the horizontal diameter (height R); a radius is drawn from the centre to the lower right; a vertical arrow on the right is labelled g.

Part B A bubble of radius $r = 5.00\ \mathrm{cm}$ , containing a diatomic ideal gas, has the soap film of thickness $h = 10.0\ \mathrm{\mu m}$ and is placed in vacuum. The soap film has the surface tension $\sigma = 4.00 \cdot 10^{-2}\ \mathrm{\frac{N}{m}}$ and the density $\rho = 1.10\ \mathrm{\frac{g}{cm^3}}$ . 1) Find formula for the molar heat capacity of the gas in the bubble for such a process when the gas is heated so slowly that the bubble remains in a mechanical equilibrium and evaluate it; 2) Find formula for the frequency $\omega$ of the small radial oscillations of the bubble and evaluate it under the assumption that the heat capacity of the soap film is much greater than the heat capacity of the gas in the bubble. Assume that the thermal equilibrium inside the bubble is reached much faster than the period of oscillations.

Hint: Laplace showed that there is pressure difference between inside and outside of a curved surface, caused by surface tension of the interface between liquid and gas, so that $\Delta p = \frac{2\sigma}{r}$ . [3 т.]

Part C Initially, a switch $S$ is unshorted in the circuit shown in the figure on the right, a capacitor of capacitance $2C$ carries the electric charge $q_0$ , a capacitor of capacitance $C$ is uncharged, and there are no electric currents in both coils of inductance $L$ and $2L$ , respectively. The capacitor starts to discharge and at the moment when the current in the coils reaches its maximum value, the switch $S$ is instantly shorted. Find the maximum current $I_{\max}$ through the switch $S$ thereafter. [3 т.]

Circuit: capacitor 2C (upper plate +, lower plate −) in the upper left branch and capacitor C in the lower left branch; coil L in the upper right branch and coil 2L in the lower right branch; a switch S connects the junction between the two capacitors to the junction between the two coils.

Отговори

Решение

Part A

Consider the forces acting on the puck and the cylinder and depicted in the figure on the right. The puck is subject to the gravity force $mg$ and the reaction force from the cylinder $N$ . The cylinder is subject to the gravity force $Mg$ , the reaction force from the plane $N_1$ , the friction force $F_{fr}$ and the pressure force from the puck $N' = -N$ . The idea is to write the horizontal projections of the equations of motion. It is written for the puck as follows

ma_x = N\sin\alpha, \qquad \text{(A.1)}

where $a_x$ is the horizontal projection of the puck acceleration.

For the cylinder the equation of motion with the acceleration $w$ is found as

Mw = N\sin\alpha - F_{fr}. \qquad \text{(A.2)}

Since the cylinder moves along the plane without sliding its angular acceleration is obtained as

\varepsilon = w/R \qquad \text{(A.3)}

Force diagram for Part A: a circle (the cylinder) on a horizontal line; the puck on the lower left inner surface with forces N (towards the centre), N′ (outwards) and mg (down); a dashed line from the centre to the puck makes angle α with the vertical; at the centre Mg points down; at the contact point N1 points up and F_fr points right along the plane.

Then the equation of rotational motion around the center of mass of the cylinder takes the form

I\varepsilon = F_{fr}R, \qquad \text{(A.4)}

where the inertia moment of the hollow cylinder is given by

I = MR^2. \qquad \text{(A.5)}

Solving (A.2)-(A.5) yields

2Mw = N\sin\alpha. \qquad \text{(A.6)}

From equations (A.1) and (A.6) it is easily concluded that

ma_x = 2Mw. \qquad \text{(A.7)}

Since the initial velocities of the puck and of the cylinder are both equal to zero, then, it follows from (A.7) after integrating that

mu = 2Mv. \qquad \text{(A.8)}

It is obvious that the conservation law for the system is written as

mgR = \frac{mu^2}{2} + \frac{Mv^2}{2} + \frac{I\omega^2}{2}, \qquad \text{(A.9)}

where the angular velocity of the cylinder is found to be

\omega = \frac{v}{R}, \qquad \text{(A.10)}

since it does not slide over the plane.

Solving (A.8)-(A.10) results in velocities at the lowest point of the puck trajectory written as

u = 2\sqrt{\frac{MgR}{(2M+m)}}, \qquad \text{(A.12)}

v = \frac{m}{M}\sqrt{\frac{MgR}{(2M+m)}}. \qquad \text{(A.13)}

In the reference frame sliding progressively along with the cylinder axis, the puck moves in a circle of radius $R$ and, at the lowest point of its trajectory, have the velocity

v_{rel} = u + v \qquad \text{(A.14)}

and the acceleration

a_{\mathrm{rel}} = \frac{v_{\mathrm{rel}}^2}{R}. \qquad \text{(A.15)}

At the lowest point of the puck trajectory the acceleration of the cylinder axis is equal to zero, therefore, the puck acceleration in the laboratory reference frame is also given by (A.15).

F - mg = \frac{mv_{rel}^2}{R}. \qquad \text{(A.16)}

then the interaction force between the puck and the cylinder is finally found as

F = 3mg\left(1 + \frac{m}{3M}\right). \qquad \text{(A.17)}

Part B

According to the first law of thermodynamics, the amount of heat transmitted $\delta Q$ to the gas in the bubble is found as

\delta Q = vC_V dT + pdV, \qquad \text{(B.1)}

where the molar heat capacity at arbitrary process is as follows

C = \frac{1}{v}\frac{\delta Q}{dT} = C_V + \frac{p}{v}\frac{dV}{dT}. \qquad \text{(B.2)}

Here $C_V$ stands for the molar heat capacity of the gas at constant volume, $p$ designates its pressure, $v$ is the total amount of moles of gas in the bubble, $V$ and $T$ denote the volume and temperature of the gas, respectively.

Evaluate the derivative standing on the right hand side of (B.2). According to the Laplace formula, the gas pressure inside the bubble is defined by

p = \frac{4\sigma}{r}, \qquad \text{(B.3)}

thus, the equation of any equilibrium process with the gas in the bubble is a polytrope of the form

p^3V = \mathrm{const}. \qquad \text{(B.4)}

The equation of state of an ideal gas has the form

pV = vRT, \qquad \text{(B.5)}

and hence equation (B.4) can be rewritten as

T^3V^{-2} = \mathrm{const}. \qquad \text{(B.6)}

Differentiating (B.6) the derivative with respect to temperature sought is found as

\frac{dV}{dT} = \frac{3V}{2T}. \qquad \text{(B.7)}

Taking into account that the molar heat capacity of a diatomic gas at constant volume is

C_V = \frac{5}{2}R, \qquad \text{(B.8)}

and using (B.5) it is finally obtained that

C = C_V + \frac{3}{2}R = 4R = 33.2\ \mathrm{\frac{J}{mole \cdot K}}. \qquad \text{(B.9)}

Since the heat capacity of the gas is much smaller than the heat capacity of the soap film, and there is heat exchange between them, the gas can be considered as isothermal since the soap film plays the role of thermostat. Consider the fragment of soap film, limited by the angle $\alpha$ as shown in the figure. It's area is found as

S = \pi(\alpha r)^2. \qquad \text{(B.10)}

and the corresponding mass is obtained as

m = \rho Sh. \qquad \text{(B.11)}

Sphere (the bubble) with a dashed equator; a small elliptical fragment of soap film on the surface subtends angle α at the centre; surface tension forces F_ST act along the fragment's edge and the pressure force F_p points radially outwards.

Let $x$ be an increase in the radius of the bubble, then the Newton second law for the fragment of the soap film mentioned above takes the form

m\ddot{x} = p'S' - F_{surf}, \qquad \text{(B.12)}

where $F_{surf}$ denotes the projection of the resultant surface tension force acting in the radial direction, $p'$ stands for the gas pressure beneath the surface of the soap film and

S' = S\left(1 + 2\frac{x}{r}\right).

$F_{surf}$ is easily found as

F_{surf} = F_{ST}\alpha = \sigma \cdot 2 \cdot 2\pi[(r+x)\alpha] \cdot \alpha. \qquad \text{(B.13)}

Since the gaseous process can be considered isothermal, it is written that

p'V' = pV. \qquad \text{(B.14)}

Assuming that the volume increase is quite small, (B.14) yields

p' = p\frac{1}{\left(1+\frac{x}{r}\right)^3} \approx p\frac{1}{\left(1+\frac{3x}{r}\right)} \approx p\left(1 - \frac{3x}{r}\right). \qquad \text{(B.15)}

Thus, from (B.10) - (B.16) and (B.3) the equation of small oscillations of the soap film is derived as

\rho h\ddot{x} = -\frac{8\sigma}{r^2}x \qquad \text{(B.16)}

with the frequency

\omega = \sqrt{\frac{8\sigma}{\rho h r^2}} = 108\ \mathrm{s^{-1}}. \qquad \text{(B.17)}

Part C

The problem can be solved in different ways. Herein several possible solutions are considered.

Method 1. Direct approach

At the moment when the current in the coils is a maximum, the total voltage across the coils is equal to zero, so the capacitor voltages must be equal in magnitude and opposite in polarity. Let $U$ be a voltage on the capacitors at the time moment just mentioned and $I_0$ be that maximum current. According to the law of charge conservation

q_0 = 2CU + CU, \qquad \text{(C1.1)}

thus,

U = \frac{q_0}{3C}. \qquad \text{(C1.2)}

Then, from the energy conservation law

\frac{q_0^2}{2 \cdot 2C} = \frac{LI_0^2}{2} + \frac{2LI_0^2}{2} + \frac{CU^2}{2} + \frac{2CU^2}{2} \qquad \text{(C1.3)}

the maximum current is found as

I_0 = \frac{q_0}{3\sqrt{2LC}}. \qquad \text{(C1.4)}

After the key $K$ is shortened there will be independent oscillations in both circuits with the frequency

\omega = \frac{1}{\sqrt{2LC}}, \qquad \text{(C1.5)}

and their amplitudes are obtained from the corresponding energy conservation laws written as

\frac{2CU^2}{2} + \frac{LI_0^2}{2} = \frac{LJ_1^2}{2}, \qquad \text{(C1.6)}

\frac{CU^2}{2} + \frac{2LI_0^2}{2} = \frac{2LJ_2^2}{2}. \qquad \text{(C1.7)}

The Part C circuit with the switch closed: upper loop with capacitor 2C (+ on top, − below) and coil L carrying loop current I1, lower loop with capacitor C and coil 2L carrying loop current I2; curved arrows show the positive directions of I1 and I2.

Hence, the corresponding amplitudes are found as

J_1 = \sqrt{5}I_0, \qquad \text{(C1.8)}

J_2 = \sqrt{2}I_0. \qquad \text{(C1.9)}

Choose the positive directions of the currents in the circuits as shown in the figure on the right. Then, the current flowing through the key is written as follows

I = I_1 - I_2. \qquad \text{(C1.10)}

The currents depend on time as

I_1(t) = A\cos\omega t + B\sin\omega t, \qquad \text{(C1.11)}

I_2(t) = D\cos\omega t + F\sin\omega t, \qquad \text{(C1.12)}

The constants $A, B, D, F$ can be determined from the initial values of the currents and their amplitudes by putting down the following set of equations

I_1(0) = A = I_0, \qquad \text{(C1.13)}

A^2 + B^2 = J_1^2, \qquad \text{(C1.14)}

I_2(0) = D = I_0, \qquad \text{(C1.15)}

D^2 + F^2 = J_2^2. \qquad \text{(C1.16)}

Solving (C1.13)-(C1.16) it is found that

B = 2I_0, \qquad \text{(C1.17)}

F = -I_0, \qquad \text{(C1.18)}

The sign in $F$ is chosen negative, since at the time moment of the key shortening the current in the coil $2L$ decreases.

Thus, the dependence of the currents on time takes the following form

I_1(t) = I_0(\cos\omega t + 2\sin\omega t), \qquad \text{(C1.19)}

I_2(t) = I_0(\cos\omega t - \sin\omega t). \qquad \text{(C1.20)}

In accordance with (C.10), the current in the key is dependent on time according to

I(t) = I_1(t) - I_2(t) = 3I_0\sin\omega t. \qquad \text{(C1.21)}

Hence, the amplitude of the current in the key is obtained as

I_{\max} = 3I_0 = \omega q_0 = \frac{q_0}{\sqrt{2LC}}. \qquad \text{(C1.22)}

Method 2. Vector diagram

Instead of determining the coefficients $A, B, D, F$ the vector diagram shown in the figure on the right can be used. The segment $AC$ represents the current sought and its projection on the current axis is zero at the time of the key shortening. The current $I_1$ in the coil of inductance $L$ grows at the same time moment because the capacitor $2C$ continues to discharge, thus, this current is depicted in the figure by the segment $OA$ . The current $I_2$ in the coil of inductance $2L$ decreases at the time of the key shortening since it continues to charge the capacitor $2C$ , that is why this current is depicted in the figure by the segment $OC$ .

It is known for above that $OB = I_0, OA = \sqrt{5}I_0, OC = \sqrt{2}I_0$ . Hence, it is found from the Pythagorean theorem that

AB = \sqrt{OA^2 - OB^2} = 2I_0, \qquad \text{(C2.1)}

BC = \sqrt{OC^2 - OB^2} = I_0, \qquad \text{(C2.2)}

Thus, the current sought is found as

I_{\max} = AC = AB + BC = 3I_0 = \omega q_0 = \frac{q_0}{\sqrt{2LC}}. \qquad \text{(C2.3)}

Vector diagram: horizontal current axis from point O labelled I; segment OB labelled I0 on the axis; vector OC labelled I2 pointing up-right to C; vector OA labelled I1 pointing down to A; vertical segment AC through B labelled Imax.

Method 3. Heuristic approach

It is clear that the current through the key performs harmonic oscillations with the frequency

\omega = \frac{1}{\sqrt{2LC}}. \qquad \text{(C3.1)}

and it is equal to zero at the time of the key shortening, i.e.

I(t) = I_{\max}\sin\omega t. \qquad \text{(C3.2)}

Since the current is equal to zero at the time of the key shortening, then the current amplitude is equal to the current derivative at this time moment divided by the oscillation frequency. Let us find that current derivative. Let the capacitor of capacitance $2C$ have the charge $q_1$ . Then the charge on the capacitor of capacitance $C$ is found from the charge conservation law as

q_2 = q_0 - q_1. \qquad \text{(C3.3)}

After shortening the key the rate of current change in the coil of inductance $L$ is obtained as

\dot{I}_1 = \frac{q_1}{2LC}, \qquad \text{(C3.4)}

whereas in the coil of inductance $2L$ it is equal to

\dot{I}_2 = -\frac{q_0 - q_1}{2LC}. \qquad \text{(C3.5)}

Since the voltage polarity on the capacitors are opposite, then the current derivative with respect to time finally takes the form

\dot{I} = \dot{I}_1 - \dot{I}_2 = \frac{q_0}{2LC} = \omega^2 q_0. \qquad \text{(C3.6)}

Note that this derivative is independent of the time of the key shortening!

Hence, the maximum current is found as

I_{\max} = \frac{\dot{I}}{\omega} = \omega q_0 = \frac{q_0}{\sqrt{2LC}}, \qquad \text{(C3.7)}

and it is independent of the time of the key shortening!

Оригинал в Архива: 1.pdf · официални решения: 1_sol.pdf