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On the competition problems of IPhO 39 in Vietnam

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2009 Eur. J. Phys. 30 S105
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IOP PUBLISHING

EUROPEAN JOURNAL OF PHYSICS


5 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.
2. 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

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Figure 1. A water-powered rice-pounding mortar.

a =20cm


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 2.
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.

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

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µ
2.7 N.m

A

23.6o

O

20.6o B

-4.6 cos

30o

0 N.m

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
2
ρ
α0 πbh2 ρ dt
should be at least dm = − 2bh
; dm is maximum at t = 0
dα = 2 2τ
cos 2πt
τ
sin2 β
sin2 β
2

2

ρ α0
ρ α0
and equals dm0 = πbh


R2
V

V1

D1

D2

V2

Figure 4. 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
0 ◦ 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 4 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.

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Vsamp[mV]

520

510

500

490

0

150

t [s]

300

450

Figure 5. Voltage of the sample as a function of time.

V [mV]

-4


according to the formula
T (t) =

T (t) =
where

k

T (0) e− C t ,
k

T (0) is the temperature difference at t = 0.

(2)


On the competition problems of IPhO 39 in Vietnam

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4

P [mW]

3

2

1


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 7. 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
3. Results

The results of the competition for all the 376 students are presented in the charts in figure 8. The
mean value and standard deviation of the scores are given in table 1. 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


6

7

8

9

100

Number of contestants

Number of contestants

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10

90
80
70
60
50
40
30
20
10
0

0

6

7

8

9

12

14

16

18

10

6

7

8

9

35
30
25
20

4

Points
Number of contestants

Number of contestants

Points

10

20

Points
Total score

30
25
20
15
10
5
0
0

2

4

6

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 2. 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

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Table 2. Minima for awards.

Gold medal
Silver medal
Bronze medal
Honourable mention

33 points
26 points
21 points
14 points

•
•
•
•
•