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Home Search Collections Journals About Contact us My IOPscience On the competition problems of IPhO 39 in Vietnam This content has been downloaded from IOPscience Please scroll down to see the full text 2009 Eur J Phys 30 S105 (http://iopscience.iop.org/0143-0807/30/6/S02) View the table of contents for this issue, or go to the journal homepage for more Download details: IP Address: 130.39.62.90 This content was downloaded on 28/08/2014 at 12:58 Please note that terms and conditions apply IOP PUBLISHING EUROPEAN JOURNAL OF PHYSICS doi:10.1088/0143-0807/30/6/S02 Eur J Phys 30 (2009) S105–S113 On the competition problems of IPhO 39 in Vietnam Nguyen The Khoi1 , Pham Quy Tu1 and Dam Trung Don2 Hanoi National University of Education, Vietnam Hanoi University of Natural Sciences, Vietnam Received June 2009, in final form 21 September 2009 Published 19 October 2009 Online at stacks.iop.org/EJP/30/S105 Abstract We report on the competition problems given at the 39th International Physics Olympiad held in Hanoi, Vietnam, which consist of three theoretical and one experimental problems The abbreviated solutions are presented The distribution of marks for the problems is also shown (Some figures in this article are in colour only in the electronic version) Introduction The 39th International Physics Olympiad (IPhO 39) took place in Hanoi, Vietnam from 20–29 July 2008 Leaders and students from 82 countries and territories participated According to the Statutes of the IPhOs [1] the theoretical examination consists of three problems of h duration The experimental examination consists of one or two problems of h duration The knowledge required to solve the problems is included in the syllabus [2] During the preparation of the problems, the academic committee also paid attention to some other aspects, namely, the relation of physics to phenomena observed in nature, in daily life, the application of physics, pollution and the protection of the environment, and energy and the use of renewable sources of energy, facts related to the host country Vietnam Problems The theoretical part of the competition consists of three problems The first one entitled Water-powered rice pounding mortar will be presented in detail in subsection 2.1 The second problem deals with the Cherenkov effect and its application in ring imaging counters The third problem studies the change of pressure in an atmosphere where the temperature depends on the altitude, and the change of temperature of an adiabatic air parcel moving in this atmosphere The obtained results are used to estimate carbon monoxide pollution in Hanoi city These two problems can be found on the website of the IPhO: www.jyu.fi/ipho/ The experimental problem Differential thermometric method will be presented in subsection 2.2 0143-0807/09/060105+09$30.00 c 2009 IOP Publishing Ltd Printed in the UK S105 N T Khoi et al S106 Figure A water-powered rice-pounding mortar a =20cm Bucket T Lever L = 74 cm cm N Pestle G =300 h= 12 cm b =15cm Mortar Figure Design and dimensions of the rice-pounding mortar 2.1 Theoretical problem no 1: water-powered rice-pounding mortar A rice-pounding mortar is shown in figure It consists of the following parts, as seen in figure 2: the mortar, basically a wooden container for rice and the lever, which is a tree trunk with one larger and one smaller end The lever can rotate around a horizontal axis A pestle is attached perpendicularly to the lever at the smaller end The length of the pestle is such that it touches the rice in the mortar when the lever lies horizontally The larger end of the lever is carved hollow to form a bucket The shape of the bucket is crucial for the mortar’s operation Consider a water-powered rice-pounding mortar with the parameters shown in figure The mass of the lever (including the pestle but without water) is M = 30 kg The centre of mass of the lever is G The lever rotates around the axis T (projected onto the point T on the figure) The moment of inertia of the lever around T is I = 12 kg m2 When there is water in the bucket, the mass of water is denoted as m, and the centre of mass of the water body is denoted as N The tilt angle of the lever with respect to the horizontal axis is α Neglect friction at the rotation axis and the force due to water falling onto the bucket In this problem, we make an approximation that the water surface is always horizontal (1) At the beginning, the bucket is empty, and the lever (TG) lies horizontally Then water flows into the bucket until the lever starts rotating The amount of water in the bucket at this moment is m = 1.0 kg 1.1 Determine TG On the competition problems of IPhO 39 in Vietnam S107 1.2 Water starts flowing out of the bucket when the angle between the lever and the horizontal axis reaches α1 The bucket is completely empty when this angle is α2 Determine α1 and α2 1.3 Let μ(α) be the total torque (relative to the axis T) which comes from the weight of the lever and the water in the bucket μ(α) is zero when α = β Determine β and the mass m1 of water in the bucket at this instant (2) Let water flow into the bucket with a flow rate , which is constant and small The amount of water flowing into the bucket, when the lever is in motion, is negligible 2.1 Sketch a graph of the torque μ as a function of the angle α, μ(α), during one operation cycle 2.2 From the graph discuss and give the geometric interpretation of the value of the total work Wtotal produced by μ(α) and the work Wpounding that is transferred from the pestle to the rice 2.3 From the graph estimate α0 and Wpounding (assume that the kinetic energy of water flowing out of the bucket is negligible) One may replace curved lines by zigzag lines, if it simplifies the calculation (3) Let water flow into the bucket with a constant rate , but one cannot neglect the amount of water flowing into the bucket during the motion of the lever 3.1 Assume that the bucket is always overflown with water 3.1.1 To which kind of equilibrium does the position α = β of the lever belong? 3.1.2 Find the analytic form of the torque μ(α) as a function of α when α = β + α, and α is small 3.1.3 Write down the equation of motion of the lever, which moves with zero initial velocity from the position α = β + α ( α is small) Show that the motion is, with good accuracy, a harmonic oscillation Compute the period τ 3.2 At a given , the bucket is overflown with water at all times only if the lever moves sufficiently slowly Determine the minimal value of (in kg/s) so that the lever can make a harmonic oscillation with amplitude 1◦ 3.3 If is too large the mortar cannot operate Assuming that the motion of the lever is that of a harmonic oscillator, estimate the minimal flow rate for the rice-pounding mortar not to work Solutions (1) 1.1 TG = 0.016 m 1.2 At the tilt angle α1 , the water level is at the edge of the bucket and the water volume is 10−3 m3 From the geometry of the bucket, we find α1 = 20.6◦ and α2 = 30◦ 1.3 m = 0.61 kg and β = 23.6◦ (2) 2.1 The graph of μ(α) is shown in figure 2.2 Wtotal = the area enclosed by the curve (OABCDFO) Wpounding = the area of (OEDFO) = gM × TG × sin α0 = 4.6 sin α0 2.3 Approximating (OABO) by a triangle and (BEDCB) by a trapezoid, we obtain α0 = 34.7◦ and Wpounding ≈ 2.6 J (3) 3.1 3.1.1 α = β is a stable equilibrium of the lever 3.1.2 By calculating the mass of water in the bucket when the lever tilts with angle α, we find that when α increases from β to β + α, the mass of water increases by N T Khoi et al S108 µ 2.7 N.m A 23.6o O 20.6o B -4.6 cos 30o N.m E D -4.0 N.m -4.6 N.m F C Figure Graph of μ(α) 2 ρ ρ m = − 2bh α ≈ − 2bh α The torque μ acting on the lever when the tilt sin2 α sin2 β is β + α equals the torque due to m, and μ = m × g × TN × cos(β + α) or μ = −47.2 × α N m ≈ −47 × α N m 3.1.3 The equation of motion of the lever is μ = I ddtα2 , where μ = − 47 × α, α = β + α and I = I (α) ≈ I (β); I = 12 + 0.6 × 0.782 = 12.36 ≈ 12.4 kg m2 We have −47 × α = 12.4 × ddt 2α That is the equation for a harmonic oscillator with period τ = 2π 12.4 47 = 3.227 ≈ 3.2 s 3.2 Assume that the lever oscillates harmonically with amplitude α0 around α = β The equation of motion is α = − α0 sin(2π t/τ ), therefore d( α) = dα = − α0 (2π/τ ) cos(2π t/τ ) dt For the bucket to be overflown, during the time dt the amount of water falling to the bucket ρ α0 πbh2 ρ dt should be at least dm = − 2bh ; dm is maximum at t = dα = 2τ cos 2πt τ sin2 β sin2 β 2 ρ α0 ρ α0 and equals dm0 = πbh dt Because dm0 = dt, then = πbh The condition τ sin2 β τ sin2 β for the lever to have a harmonic oscillation with amplitude α0 = 1◦ is with πbh2 ρ2π −1 −1 ≈ 0.23 kg s = 360τ sin2 β = 0.2309 kg s 3.3 If the bucket remains overflown when the tilt decreases to 20.6◦ , then the amount of water in the bucket should reach kg at this time, and the lever oscillates harmonically with amplitude equal to β − α1 = 23.6◦ − 20.6◦ = 3◦ The flow should exceed , therefore = × 0.23 ≈ 0.7 kgs−1 This is the minimal flow rate for the rice-pounding mortar not to work Discussion Neglecting the angular momentum variation of the lever due to the collision of water falling onto the bucket simplifies the solution If one takes it into account, the shape of the curve μ(α) is nearly unchanged; only the maximal angle α0 increases and therefore Wpounding may also increase On the competition problems of IPhO 39 in Vietnam R1 E S109 R2 V V1 D1 D2 V2 Figure Schematic circuit diagram of the experimental setup 2.2 Experimental problem; differential thermometric method In this experiment, forward-biased silicon diodes are used as sensors to measure temperature The voltage drop across a diode at constant bias current depends linearly on the diode’s temperature T as V (T ) = V (T0 ) − α(T − T0 ), where ◦ C < T < 100 ◦ C, T0 is the room temperature and α = (2.00 ± 0.03) mV ◦ C−1 If two diodes D1 and D2 are connected in an electric circuit as shown in figure and placed at different temperatures T1 and T2 respectively, then the voltage V , called the differential voltage, is V = V2 (T2 ) − V1 (T1 ) = V (T0 ) − α(T2 − T1 ) By measuring V , we can determine T = T2 − T1 This method is called the differential thermometric method 2.2.1 Differential thermometric method We use the differential thermometric method to determine the temperature of solidification Ts of a crystalline substance Two small identical metallic dishes, each with a diode temperature sensor fixed to it, are put on a steel plate The sample dish contains a small amount (about 10 mg) of the substance to be studied The reference dish is empty The steel plate with the two dishes is first heated by a halogen lamp to a temperature higher than Ts Then the lamp is switched off, and the plate is left to cool down While there is no phase change in the substance, the temperatures Tsamp of the sample dish and Tref of the reference dish vary at nearly the same rate, and thus T = Tref − Tsamp varies slowly with Tsamp During the phase change (the solidification), Tsamp does not vary and equals Ts , while Tref steadily varies, so that T varies quickly The plot of T versus Tsamp shows an abrupt change The value of Tsamp corresponding to the abrupt change of T is indeed Ts The student is asked to monitor the change in Vsamp and V with time t and to note the values of Vsamp and V every 10 s–20 s in a table He or she plots Vsamp versus t and V versus Vsamp Examples of these two graphs are given in figures and From figure 5, the student deduces Ts by the traditional method From figure 6, he/she deduces Ts by the differential thermometric method 2.2.2 Task Finding the temperature of solidification of a crystalline substance 2.2.3 Task Determining the maximum efficiency of a solar cell under illumination of an incandescent lamp P (1) The efficiency of a solar cell is determined as η = E×S , where P is the electric power cell the solar cell provides for the load, E is the irradiance and Scell is the area of the solar cell To measure the irradiance, a radiation detector is used, which consists of a hollow cone made of copper, the inner surface of which is blackened with soot The measuring diode D2 is fixed to the cone The reference diode D1 is placed on the inner side of the wall of the box containing the detector; its temperature equals that of the surroundings N T Khoi et al S110 Vsamp[mV] 520 510 500 490 150 t [s] 300 450 Figure Voltage of the sample as a function of time V [mV] -4 -5 -6 -7 490 500 510 520 Vsamp[mV] Figure Voltage difference in relation to the voltage on the sample T0 When the detector receives energy from radiation, it heats up At the same time, the detector loses its heat by several mechanisms, such as thermal conduction, convection, radiation etc The detector is placed under a lamp, which serves as a light source The variation of the difference between the temperature of the detector and that of the surroundings T = T − T0 with time t from the moment the detector begins to receive the light with constant irradiation is given by − e− C t , (1) k where C is the heat capacity of the detector, whose value is given, is the radiant power which is related to the area S of the illuminated surface by E = /S Assume that at t = 0, T = 0, and the heat-missing rate is k T , where k is a constant When the radiation is switched off, the temperature difference T varies with time according to the formula T (t) = T (t) = where k T (0) e− C t , k T (0) is the temperature difference at t = (2) On the competition problems of IPhO 39 in Vietnam S111 P [mW] 3 11 13 I [mA] Figure Electric power in the load as a function of the current through the cell (2) The student has to place the detector under the lamp and switch the lamp on to illuminate the detector Every 10–15 s, he/she writes down a value of V in a table After min, he/she switches the lamp off The lamp is then moved away from the detector Every 10–15 s, the student writes down a value of V in a table With the data obtained when the detector is illuminated, the student is asked to plot a corresponding graph, with variables x and y chosen appropriately, in order to prove that equation (1) is satisfied He/she has also to plot the data when the detector ceases to be illuminated in order to prove that equation (2) is satisfied The student has to choose two sets of variables x and y for the two cases, so that the plots are linear From the plots, the student can determine the value of k and and calculate the value of E (3) The student now puts the solar cell in the place where the radiation detector was, and connects the solar cell to an appropriate electric circuit comprising multimetres and a variable resistor He/she is asked to change the value of the load of the cell by rotating the knob of the variable resistor, to note the values of current I and voltage V at each position of the knob and to plot a graph of the electric power in the load, that is I × V , as a function of the current I through the cell From the graph, the student determines the maximum power Pmax that the solar cell can provide for the load and estimate its error Then he/she calculates the maximum efficiency of the solar cell and its error We give an example of the solution in figure The maximal power of the solar cell calculated from the graph is Pmax = 3.7 ± 0.2 mW Pmax = 0.058 ± 0.008 The maximal efficiency of the solar cell is ηmax = E×S cell Results The results of the competition for all the 376 students are presented in the charts in figure The mean value and standard deviation of the scores are given in table As stated in the Statutes, from the grading results the organizers establish minima (expressed in points) according to the following rules: (a) (b) (c) (d) A gold medal should be awarded to 6% of the contestants Gold or silver medals should be awarded to 18% of the contestants Gold, silver or bronze medals should be awarded to 36% of the contestants An Olympic medal or honourable mention should be awarded to 60% of the contestants N T Khoi et al Theory Theory 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0 100 Number of contestants Number of contestants S112 10 90 80 70 60 50 40 30 20 10 0 110 100 90 80 70 60 50 40 30 20 10 12 14 16 18 10 35 30 25 20 15 10 10 Points Number of contestants Experiment Theory 120 Points Number of contestants Number of contestants Points 10 20 Points Total score 30 25 20 15 10 0 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 Points Figure Distribution of the scores Table Mean value and standard deviation of the scores Problem Mean value Standard deviation Theoretical problem Theoretical problem Theoretical problem Experimental problem Total score 2.21 3.40 2.55 11.96 20.12 1.53 2.77 2.72 4.83 9.68 The corresponding minima should be expressed as integers by rounding off to the nearest lower integers The minima for awarding the medals and the honourable mention were established for IPhO 39, as given in table According to these limits, 46 gold medals, 47 silver medals, 78 bronze medals and 87 honourable mentions were awarded Special awards were given to On the competition problems of IPhO 39 in Vietnam S113 Table Minima for awards Gold medal Silver medal Bronze medal Honourable mention 33 points 26 points 21 points 14 points • • • • • Longzhi Tan (China) for the best total score (44.6 points) Longzhi Tan (China) for the best score in the theoretical part (26.1 points) Yi-Shu Wei (Taiwan) for the best score in the experimental part (20 points) Andrada Ianus (Romania) for the best score among female participants Efrain Alfonso P´erez Argandona (Chile) for the best participant among the countries that first joined the IPhO in 2008 (Gorzkowski Prize) • Huynh Minh Toan (Vietnam) for the best participant from the host country In conclusion, we have presented the competition problems and the main results of IPhO 39, Vietnam References [1] Statutes of the International Physics Olympiads http://www.jyu.fi/ipho/ [2] The Syllabus (Appendix to the Statutes of the International Physics Olympiads) http://www.jyu.fi/ipho/ ... examination consists of three problems of h duration The experimental examination consists of one or two problems of h duration The knowledge required to solve the problems is included in the syllabus... friction at the rotation axis and the force due to water falling onto the bucket In this problem, we make an approximation that the water surface is always horizontal (1) At the beginning, the. .. report on the competition problems given at the 39th International Physics Olympiad held in Hanoi, Vietnam, which consist of three theoretical and one experimental problems The abbreviated solutions