2014 Ipho Solution To Theory Problem 1 6ri2b

Theoretical competition. Tuesday, 15 July 2014

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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 𝑚𝑔 and the reaction force from the cylinder 𝑁. The cylinder is subject to the gravity force 𝑀𝑔, the reaction force from the plane 𝑁1 , the friction force 𝐹𝑓𝑟 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 𝑚𝑎𝑥 = 𝑁 sin 𝛼, (A.1) where 𝑎𝑥 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 𝛼 − 𝐹𝑓𝑟 . (A.2) Since the cylinder moves along the plane without sliding its angular acceleration is obtained as 𝜀 = 𝑤/𝑅 (A.3) Then the equation of rotational motion around the center of mass of the cylinder takes the form 𝐼𝜀 = 𝐹𝑓𝑟 𝑅, (A.4) where the inertia moment of the hollow cylinder is given by 𝐼 = 𝑀𝑅 2 . (A.5) Solving (A.2)-(A.5) yields 2𝑀𝑤 = 𝑁 sin 𝛼. (A.6) From equations (A.1) and (A.6) it is easily concluded that 𝑚𝑎𝑥 = 2𝑀𝑤. (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. (A.8) It is obvious that the conservation law for the system is written as 𝑚𝑢2

𝑀𝑣 2

𝐼𝜔 2

𝑚𝑔𝑅 = 2 + 2 + 2 , (A.9) where the angular velocity of the cylinder is found to be 𝑣 𝜔 = 𝑅, (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

𝑀𝑔𝑅

𝑚

𝑣=𝑀

,

(A.12)

(2𝑀+𝑚 ) 𝑀𝑔𝑅

(2𝑀+𝑚 )

.

(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 𝑣𝑟𝑒𝑙 = 𝑢 + 𝑣 (A.14) and the acceleration 𝑣2

𝑎rel = rel . (A.15) 𝑅 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). 𝑚 𝑣2

𝐹 − 𝑚𝑔 = 𝑅𝑟𝑒𝑙 . then the interaction force between the puck and the cylinder is finally found as 𝑚 𝐹 = 3𝑚𝑔 1 + 3𝑀 .

(A.16) (A.17)

Theoretical competition. Tuesday, 15 July 2014

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Part B 1) According to the first law of thermodynamics, the amount of heat transmitted 𝛿𝑄 to the gas in the bubble is found as 𝛿𝑄 = 𝑣𝐶𝑉 𝑑𝑇 + 𝑝𝑑𝑉, (B.1) where the molar heat capacity at arbitrary process is as follows 1 𝛿𝑄 𝑝 𝑑𝑉 𝐶 = 𝑣 𝑑𝑇 = 𝐶𝑉 + 𝑣 𝑑𝑇 . (B.2) Here 𝐶𝑉 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 4𝜍 𝑝= 𝑟, (B.3) thus, the equation of any equilibrium process with the gas in the bubble is a polytrope of the form 𝑝3 𝑉 = const. (B.4) The equation of state of an ideal gas has the form 𝑝𝑉 = 𝑣𝑅𝑇, (B.5) and hence equation (B.4) can be rewritten as 𝑇 3 𝑉 −2 = const. (B.6) Differentiating (B.6) the derivative with respect to temperature sought is found as 𝑑𝑉 3𝑉 = 2𝑇 . (B.7) 𝑑𝑇 Taking into that the molar heat capacity of a diatomic gas at constant volume is 5 𝐶𝑉 = 2 𝑅, (B.8) and using (B.5) it is finally obtained that 3 J 𝐶 = 𝐶𝑉 + 2 𝑅 = 4𝑅 = 33.2 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 𝛼 as shown in the figure. It's area is found as 𝑆 = 𝜋(𝛼𝑟)2 . (B.10) and the corresponding mass is obtained as 𝑚 = 𝜌𝑆𝑕. (B.11) 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 𝑚𝑥 = 𝑝′ 𝑆′ − 𝐹𝑠𝑢𝑟𝑓 , (B.12) where 𝐹𝑠𝑢𝑟𝑓 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 𝑥 𝑆′ = 𝑆 1 + 2 𝑟 . 𝐹𝑠𝑢𝑟𝑓 is easily found as 𝐹𝑠𝑢𝑟𝑓 = 𝐹𝑆𝑇 𝛼 = 𝜍 ∙ 2 ∙ 2𝜋[ 𝑟 + 𝑥 𝛼] ∙ 𝛼. (B.13) Since the gaseous process can be considered isothermal, it is written that 𝑝′ 𝑉 ′ = 𝑝𝑉. (B.14) Assuming that the volume increase is quite small, (B.14) yields 1 1 3𝑥 𝑝′ = 𝑝 𝑥 3 ≈ 𝑝 3𝑥 ≈ 𝑝 1 − 𝑟 . (B.15) 1+

𝑟

1+

𝑟

Thus, from (B.10) - (B.16) and (B.3) the equation of small oscillations of the soap film is derived as 8𝜍 𝜌𝑕𝑥 = − 𝑟 2 𝑥 (B.16)

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with the frequency 8𝜍

𝜔=

𝜌𝑕𝑟 2

= 108 s −1 .

(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 𝐼0 be that maximum current. According to the law of charge conservation 𝑞0 = 2𝐶𝑈 + 𝐶𝑈, (C1.1) thus, 𝑞 𝑈 = 3𝐶0 . (C1.2) Then, from the energy conservation law 𝑞 02

𝐿𝐼 2

= 20 + 2∙2𝐶 the maximum current is found as 𝑞0 𝐼0 = 3 2𝐿𝐶 .

2𝐿𝐼02 2

+

𝐶𝑈 2 2

+

2𝐶𝑈 2 2

(C1.3) (C1.4)

After the key 𝐾 is shortened there will be independent oscillations in both circuits with the frequency 1 𝜔 = 2𝐿𝐶 , (C1.5) and their amplitudes are obtained from the corresponding energy conservation laws written as 2𝐶𝑈 2 2 𝐶𝑈 2

+

𝐿𝐼02

2 2𝐿𝐼02

=

𝐿𝐽 12

,

2 2𝐿𝐽 22

(C1.6)

+ 2 = 2 . (C1.7) Hence, the corresponding amplitudes are found as 𝐽1 = 5𝐼0 , (C1.8) 𝐽2 = 2𝐼0 . (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 = 𝐼1 − 𝐼2 . (C1.10) The currents depend on time as 𝐼1 𝑡 = 𝐴 cos 𝜔𝑡 + 𝐵 sin 𝜔𝑡, (C1.11) 𝐼2 𝑡 = 𝐷 cos 𝜔𝑡 + 𝐹 sin 𝜔𝑡, (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 𝐼1 0 = 𝐴 = 𝐼0 , (C1.13) 2 2 2 𝐴 + 𝐵 = 𝐽1 , (C1.14) 𝐼2 0 = 𝐷 = 𝐼0 , (C1.15) 2 2 2 𝐷 + 𝐹 = 𝐽2 . (C1.16) Solving (C1.13)-(C1.16) it is found that 𝐵 = 2𝐼0 , (C1.17) 𝐹 = −𝐼0 , (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 𝐼1 𝑡 = 𝐼0 (cos 𝜔𝑡 + 2 sin 𝜔𝑡), (C1.19) 𝐼2 𝑡 = 𝐼0 (cos 𝜔𝑡 − sin 𝜔𝑡). (C1.20) In accordance with (C.10), the current in the key is dependent on time according to 𝐼 𝑡 = 𝐼1 𝑡 − 𝐼2 𝑡 = 3𝐼0 sin 𝜔𝑡. (C1.21) Hence, the amplitude of the current in the key is obtained as 𝑞0 𝐼max = 3𝐼0 = 𝜔𝑞0 = 2𝐿𝐶 . (C1.22) 2

Theoretical competition. Tuesday, 15 July 2014 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 𝐼1 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 𝐼2 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 = 𝐼0 , 𝑂𝐴 = 5𝐼0 , 𝑂𝐶 = 2𝐼0 . Hence, it is found from the Pythagorean theorem that 𝐴𝐵 = 𝑂𝐴2 − 𝑂𝐵 2 = 2𝐼0 , (C2.1) 2 2 𝐵𝐶 = 𝑂𝐶 − 𝑂𝐵 = 𝐼0 , (C2.2) Thus, the current sought is found as 𝑞0 𝐼max = 𝐴𝐶 = 𝐴𝐵 + 𝐵𝐶 = 3𝐼0 = 𝜔𝑞0 = 2𝐿𝐶 .

4/4

(C2.3)

Method 3. Heuristic approach It is clear that the current through the key performs harmonic oscillations with the frequency 1 𝜔 = 2𝐿𝐶 . (C3.1) and it is equal to zero at the time of the key shortening, i.e. 𝐼 𝑡 = 𝐼max sin 𝜔𝑡. (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 𝑞1 . Then the charge on the capacitor of capacitance 𝐶 is found from the charge conservation law as 𝑞2 = 𝑞0 − 𝑞1 . (C3.3) After shortening the key the rate of current change in the coil of inductance 𝐿 is obtained as 𝑞1 𝐼1 = 2𝐿𝐶 , (C3.4) whereas in the coil of inductance 2𝐿 it is equal to 𝑞 −𝑞 𝐼2 = − 02𝐿𝐶 1 . (C3.5) Since the voltage polarity on the capacitors are opposite, then the current derivative with respect to time finally takes the form 𝑞0 𝐼 = 𝐼1 − 𝐼2 = 2𝐿𝐶 = 𝜔2 𝑞0 . (C3.6) Note that this derivative is independent of the time of the key shortening! Hence, the maximum current is found as 𝐼

𝑞

0 𝐼max = 𝜔 = 𝜔𝑞0 = 2𝐿𝐶 , and it is independent of the time of the key shortening!

(C3.7)