IPhO2014-第45届国际物理奥林匹克竞赛理论试题与解答.pdf

Theoretical competition. Tuesday, 15 July 2014 1/1 Problem 1 (9 points) This problem consists of three independent parts. Part A (3 points) A small puck of mass is carefully placed onto the inner surface of the thin hollow thin cylinder of mass and of radius . Initially, the cylinder rests on the horizontal plane and the puck is located at the height above the plane as shown in the figure on the left. Find the interaction force 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 . Part B (3 points) A bubble of radius = 5.00 cm, containing a diatomic ideal gas, has the soap film of thickness = 10.0 m and is placed in vacuum. The soap film has the surface tension = 4.00 102 N m and the density = 1.10 g cm3. 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 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 = 2 . Part C (3 points) Initially, a switch is unshorted in the circuit shown in the figure on the right, a capacitor of capacitance 2 carries the electric charge 0, a capacitor of capacitance is uncharged, and there are no electric currents in both coils of inductance and 2, respectively. The capacitor starts to discharge and at the moment when the current in the coils reaches its maximum value, the switch is instantly shorted. Find the maximum current max through the switch thereafter. Theoretical competition. Tuesday, 15 July 2014 1/4 Problem 1 Solution 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 and the reaction force from the cylinder . The cylinder is subject to the gravity force , the reaction force from the plane 1, the friction force and the pressure force from the puck = . The idea is to write the horizontal projections of the equations of motion. It is written for the puck as follows = sin, (A.1) where is the horizontal projection of the puck acceleration. For the cylinder the equation of motion with the acceleration is found as = sin . (A.2) Since the cylinder moves along the plane without sliding its angular acceleration is obtained as = / (A.3) Then the equation of rotational motion around the center of mass of the cylinder takes the form = , (A.4) where the inertia moment of the hollow cylinder is given by = 2. (A.5) Solving (A.2)-(A.5) yields 2 = sin. (A.6) From equations (A.1) and (A.6) it is easily concluded that = 2. (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 = 2. (A.8) It is obvious that the conservation law for the system is written as = 2 2 + 2 2 + 2 2 , (A.9) where the angular velocity of the cylinder is found to be = , (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 = 2 (2+), (A.12) = (2+). (A.13) In the reference frame sliding progressively along with the cylinder axis, the puck moves in a circle of radius and, at the lowest point of its trajectory, have the velocity = + (A.14) and the acceleration rel= rel 2 . (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). = 2 . (A.16) then the interaction force between the puck and the cylinder is finally found as = 3 1 + 3 . (A.17) Theoretical competition. Tuesday, 15 July 2014 2/4 Part B 1) According to the first law of thermodynamics, the amount of heat transmitted to the gas in the bubble is found as = + , (B.1) where the molar heat capacity at arbitrary process is as follows = 1 = + . (B.2) Here stands for the molar heat capacity of the gas at constant volume, designates its pressure, is the total amount of moles of gas in the bubble, and 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 = 4 , (B.3) thus, the equation of any equilibrium process with the gas in the bubble is a polytrope of the form 3 = const. (B.4) The equation of state of an ideal gas has the form = , (B.5) and hence equation (B.4) can be rewritten as 32= const. (B.6) Differentiating (B.6) the derivative with respect to temperature sought is found as = 3 2. (B.7) Taking into account that the molar heat capacity of a diatomic gas at constant volume is = 5 2 , (B.8) and using (B.5) it is finally obtained that = + 3 2 = 4 = 33.2 J mole K. (B.9) 2) 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 as shown in the figure. Its area is found as = ()2. (B.10) and the corresponding mass is obtained as = . (B.11) Let 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 = , (B.12) where denotes the projection of the resultant surface tension force acting in the radial direction, stands for the gas pressure beneath the surface of the soap film and = 1 + 2 . is easily found as = = 2 2 + . (B.13) Since the gaseous process can be considered isothermal, it is written that = . (B.14) Assuming that the volume increase is quite small, (B.14) yields = 1 1+ 3 1 1+3 1 3 . (B.15) Thus, from (B.10) - (B.16) and (B.3) the equation of small oscillations of the soap film is derived as = 8 2 (B.16) Theoretical competition. Tuesday, 15 July 2014 3/4 with the frequency = 8 2 = 108 s1. (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 be a voltage on the capacitors at the time moment just mentioned and 0 be that maximum current. According to the law of charge conservation 0= 2 + , (C1.1) thus, = 0 3. (C1.2) Then, from the energy conservation law 0 2 22 = 0 2 2 + 20 2 2 + 2 2 + 22 2 (C1.3) the maximum current is found as 0= 0 3 2. (C1.4) After the key is shortened there will be independent oscillations in both circuits with the frequency = 1 2, (C1.5) and their amplitudes are obtained from the corresponding energy conservation laws written as 22 2 + 0 2 2 = 1 2 2 , (C1.6) 2 2 + 20 2 2 = 22 2 2 . (C1.7) Hence, the corresponding amplitudes are found as 1= 50, (C1.8) 2= 20. (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 = 1 2. (C1.10) The currents depend on time as 1 = cos + sin, (C1.11) 2 = cos + sin, (C1.12) The constants , can be determined from the initial values of the currents and their amplitudes by putting down the following set of equations 1 0 = = 0, (C1.13) 2+ 2= 1 2, (C1.14) 2 0 = = 0, (C1.15) 2+ 2= 2 2. (C1.16) Solving (C1.13)-(C1.16) it is found that = 20, (C1.17) = 0, (C1.18) The sign in is chosen negative, since at the time moment of the key shortening the current in the coil 2 decreases. Thus, the dependence of the currents on time takes the following form 1 = 0(cos + 2sin), (C1.19) 2 = 0(cos sin). (C1.20) In accordance with (C.10), the current in the key is dependent on time according to = 1 2 = 30sin. (C1.21) Hence, the amplitude of the current in the key is obtained as max= 30= 0= 0 2. (C1.22) Theoretical competition. Tuesday, 15 July 2014 4/4 Method 2. Vector diagram Instead of determining the coefficients , the vector diagram shown in the figure on the right can be used. The segment represents the current sought and its projection on the current axis is zero at the time of the key shortening. The current 1 in the coil of inductance grows at the same time moment because the capacitor 2 continues to discharge, thus, this current is depicted in the figure by the segment . The current 2 in the coil of inductance 2 decreases at the time of the key shortening since it continues to charge the capacitor 2, that is why this current is depicted in the figure by the segment . It is known for above that = 0, = 50, = 20. Hence, it is found from the Pythagorean theorem that = 2 2= 20, (C2.1) = 2 2= 0, (C2.2) Thus, the current sought is found as max= = + = 30= 0= 0 2. (C2.3) Method 3. Heuristic approach It is clear that the current through the key performs harmonic oscillations with the frequency = 1 2. (C3.1) and it is equal to zero at the time of the key shortening, i.e. = maxsin. (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 2 have the charge 1. Then the charge on the capacitor of capacitance is found from the charge conservation law as 2= 0 1. (C3.3) After shortening the key the rate of current change in the coil of inductance is obtained as 1= 1 2, (C3.4) whereas in the coil of inductance 2 it is equal to 2= 01 2 . (C3.5) Since the voltage polarity on the capacitors are opposite, then the current derivative with respect to time finally takes the form = 1 2= 0 2 = 20. (C3.6) Note that this derivative is independent of the time of the key shortening! Hence, the maximum current is found as max= = 0= 0 2, (C3.7) and it is independent of the time of the key shortening! Theoretical competition. Tuesday, 15 July 2014 1/3 Problem 2. Van der Waals equation of state (11 points) In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered. Part . Non-ideal gas equation of state (2 points) Taking into account the finite size of the molecules, the gaseous equation of state takes the form = , (1) where , stands for the gas pressure, its volume per mole and temperature, respectively, denotes the universal gas constant, and is a specific constant extracting some volume. A1 Estimate and express it in terms of the diameter of the molecules . (0.3 points) With account of intermolecular attraction forces, van der Waals proposed the following equation of state that neatly describes both the gaseous and liquid states of matter + 2 = . (2) where is another specific constant. At temperatures below a certain critical value the isotherm of equation (2) is well represented by a non-monotonic curve 1 shown in Figure 1 which is then called van der Waals isotherm. In the same figure curve 2 shows the isotherm of an ideal gas at the same temperature. A real isotherm differs from the van der Waals isotherm by a straight segment drawn at some constant pressure . This straight segment is located between the volumes and , and corresponds to the equilibrium of the liquid phase (indicated by ) and the gaseous phase (referred to by ). From the second law of thermodynamics J. Maxwell showed that the pressure must be chosen such that the areas I and II shown in Figure 1 must be equal. Figure 1. Van der Waals isotherm of gas/liquid (curve 1) and the isotherm of an ideal gas (curve 2). Figure 2. Several isotherms for van der Waals equation of state. With increasing temperature the straight segment on the isotherm shrinks to a single point when the temperature and the pressure reaches some values and = , respectively. The parameters and are called critical and can be measured experimentally with high degree of accuracy. A2 Express the van der Waals constants and in terms of and . (1.3 points) A3 For water = 647 K and = 2.2 107 Pa. Calculate and for water. (0.2 points) A4 Estimate the diameter of water molecules . (0.2 points) Theoretical competition. Tuesday, 15 July 2014 2/3 Part B. Properties of gas and liquid (6 points) This part of the problem deals with the properties of water in the gaseous and liquid states at temperature = 100 C. The saturated vapor pressure at this temperature is known to be = 0= 1.0 105 Pa, and the molar mass of water is = 1.8 102 kg mole. Gaseous state It is reasonable to assume that the inequality is valid for the description of water properties in a gaseous state. B1 Derive the formula for the volume and express it in terms of ,0, and . (0.8 points) Almost the same volume 0 can be approximately evaluated using the ideal gas law. B2 Evaluate in percentage the relative decrease in the gas volume due to intermolecular forces, 0 = 0 0 . (0.3 points) If the system volume is reduced below , the gas starts to condense. However, thoroughly purified gas can remain in a mechanically metastable state (called supercooled vapor) until its volume reaches a certain value min. The condition of mechanical stability of supercooled gas at constant temperature is written as: 0. 3 Find and evaluate how many times the volume of water vapor can be reduced and still remains in a metastable state. In other words, what is /min? (0.7 points) Liquid state For the van der Waals description of water in a liquid state it is reasonable to assume that the following inequality holds /2. B4 Express the volume of liquid water in terms of , and . (1.0 points) Assuming that , find the following characteristics of water. Do not be surprised if some of the data evaluated do not coincide with the well-known tabulated values! B5 Express the liquid water density in some of the terms of , and evaluate it. (0.5 points) B6 Express the volume thermal expansion coefficient = 1 in terms of , and evaluate it. (0.6 points) B7 Express the specific heat of water vaporization in terms of , and evaluate it. (1.1 points) B8 Considering the monomolecular layer of water, estimate the surface tension of water. (1.2 points) Part . Liquid-gas system (3 points) From Maxwells rule (equalities of areas, by applying trivial integration) and the van der Waals equation of state together with the approximations made in Part B, it can be shown that the saturated vapor pressure depends on temperature as follows ln= + , (3) where and are some constants, that can be expressed in terms of and as = ln 2 1; = Theoretical competition. Tuesday, 15 July 2014 3/3 W. Thomson showed that the pressure of saturated vapor depends on the curvature of the liquid surface. Consider a liquid that does not wet the material of a capillary (contact angle 180). When the capillary is immersed into the liquid, the liquid in the capillary drops to a certain level because of the surface tension (see Figure 3). 1 Find a small change in pressure of the saturated vapor over the curved surface of liquid and express it in terms of the vapor density , the liquid density , the surface tension and the radius of surface curvature . (1.3 points) Metastable states, considered in part B3, are widely used in real experimental setups, such as the cloud chamber designed for registration of elementary particles. They also occur in natural phenomena such as the formation of morning dew. Supercooled vapor is subject to condensation