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The final round of the first World Physics Olympiad held in Lombok, West Nusa Tenggara,

Indonesia: a sample of problems and solutions and student results

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IOP PUBLISHING EUROPEAN JOURNAL OF PHYSICS

Eur. J. Phys. 34 (2013) S15–S34 doi:10.1088/0143-0807/34/4/S15

The final round of the first WorldPhysics Olympiad held in Lombok,

West Nusa Tenggara, Indonesia: a

sample of problems and solutions and

student results

H J Kwee1, O Gunawan2, Y Surya1 and M Vigh3

1Physics Education Department, Surya College of Education, SURE Center Building,

Kab Tangerang, Banten 15810, Indonesia2 IBM T J Watson Research Center, Yorktown Heights, New York, NY 10598, USA3 Department of Physics of Complex Systems, Eotvos University, H-1117 Budapest,Pazmany Peter setany 1/A, Hungary

E-mail: [email protected], [email protected], [email protected] [email protected]

Received 1 March 2013, in final form 10 April 2013Published 28 May 2013Online at stacks.iop.org/EJP/34/S15

Abstract

A brief report on the final round of the first World Physics Olympiad (WoPhO)

held in Lombok, West Nusa Tenggara, Indonesia is presented. The theoreticaland experimental problems are presented and the mark distribution is discussed.

(Some figures may appear in colour only in the online journal)

1. Introduction

The final round of the first World Physics Olympiad (WoPhO) was held in Lombok, West

Nusa Tenggara, Indonesia from 28 December 2011 to 3 January 2012 and was organized by

Surya College of Education and the Indonesian Society for the Promotion of Science under

the auspices of Surya Institute.

WoPhO is a secondary school level individual physics competition initiated by one of the

authors, Yohanes Surya. It is a unique competition that lasts for a full year and consists of 

three rounds: selection, discussion and final.

The selection round is meant to provide as many opportunities as possible for students

to participate in an Olympiad-level physics competition. Problems are presented online

and any students who are eligible4 can participate. For the first WoPhO selection round,

4 Students who have not started college or become 20 years old by 30 June of the competition year.

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S16 H J Kwee et al 

Figure 1. Sketch of the rod with caster-wheels.

345 students from all over the world registered to participate. The discussion round provides

an opportunity for the participants and general public to discuss the problems and solutions

of the selection round and other physics Olympiad-level problems. The final round, which

is organized similarly to the International and Asian Physics Olympiads (IPhO and APhO)

provides the opportunity for every participant to challenge the APhO and IPhO gold medalists.

Just as in APhO and IPhO, students are given 5 h to solve the three theoretical problems and

another five for the experimental tasks. For the first WoPhO final round, 122 students from

13 countries participated, even though the distribution was not evenly spread. The majority of 

students were from the host country, Indonesia.

Another unique characteristic of WoPhO is that the problems for the final round are

not provided by the host country, but are selected from a competition (or simply called theWoPhO problem competition). Anyone can participate in this competition, in fact one of 

the winners of this first WoPhO problem competition was a high-school student and another

was a university student. Three theoretical problems were chosen from 31 entrants and two

experimental problems were chosen from four entrants. The number of entries for experimental

problems was much lower than for theoretical problems due to the amount of work needed to

prepare a good experiment and its apparatus.

Since all of the final round problems can be accessed online5 we present only one

theoretical and one experimental problem, with their solutions.

In section 2, we present one theoretical problem: ‘The motion of a rolling rod and its

solution (see footnote 5)’.

In section 3, we present one experimental problem: ‘A rotary magnetic drag system for

conductivity measurement and its solution (see footnote 5)’.In section 4, we summarize the final round results (see footnote 5).

This paper was prepared for a European Journal of Physics special issue covering physics

competitions. Other interesting papers on similar topics can be found elsewhere in the issue

[2, 3].

2. Theoretical problem 1: the motion of a rolling rod

2.1. Problem statement 

In this problem, the motion of a uniform rod (stick) with length L, with caster-wheels at

both ends, will be investigated on a flat surface. The casters at each end of the rod can spin

freely and independently (see figure 1) and have a negligible mass compared to the rod. The

friction between the rod and the caster-wheels is negligible. The diameter of the caster-wheelsis slightly larger than the diameter of the rod, but both diameters are much smaller than the

length of the rod. The gravitational acceleration is g.

(i) The rod is placed on a horizontal flat surface and pushed such that each end of the

rod has a different horizontal initial velocity (v1 and v2, pointing in the same direction)

perpendicular to the axis of the rod. The casters roll without slipping on the surface.

5 www.wopho.org/about-detail.php?id=704

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Figure 2. The initial conditions of the rod in section 3.

(a) Calculate the initial velocity v0 of the center of the rod and the initial angular velocity

ω0 of the rod using v1, v2 and L. (0.8 points)

(b) Describe the motion of the center of mass of the rod. Determine the parameter(s) of 

its orbit. (0.8 points)

(c) What should the minimum value of the coefficient of static friction μ for the casters

be to not slip on the surface? (0.6 points)

In the following sections the case of the inclined surface will be considered. The angle

between the inclined surface and the horizontal plane is α.

(ii) If  α is infinitesimally small, the motion of the rod slightly changes: the motion of the

center of mass is approximately the same as in the previous section, but with a constant

drift velocity vdrift added to the solution. Use a coordinate system as in figure 2.

(a) Calculate the magnitude and the direction of  vdrift as a function of the small α, the

initial velocities of the two ends of the rod ( v1 and v2, pointing in the same direction)

and the gravitational acceleration g! (1.9 points)

(b) Sketch the orbit of the center of mass of the rod. (0.5 points)

(iii) If  α is finite, the details of the motion of the rod change. Place the rod on the inclinedplane along the steepest line of the surface (so the rod is parallel with the inclined edges

of the plane). Consider that the initial velocity v0 of the center of mass of the rod is

perpendicular to the axis of the rod and the initial angular velocity ω0 is perpendicular to

the surface, as shown in figure 2.

(a) Calculate the time evolution of the velocity v(t ) = (v x (t ), v y(t )) of the center of 

mass of the rod in the Cartesian coordinate system shown in figure 2. (0.8 points)

(b) Depending on the magnitude of the v0 and ω0, the center of the rod can stop for a

moment during its motion. Express the condition(s) for such a behavior using the

parameters v0, ω0, g, α and L. (0.8 points)

(c) Determine the maximum displacement of the center of the rod in the direction of the

steepest line ( y-direction) as a function of v0 and ω0. (1.2 points)

(iv) Investigate another situation where the rod is placed horizontally on the inclined surface.Consider that the initial angular velocity ω0 of the rod is perpendicular to the surface but

the initial velocity of the center of the rod is zero (see figure 3).

(a) Describe the motion of the center of mass of the rod! Determine the parameter(s) of 

its orbit. (1.6 points)

(b) What should the minimum value of the coefficient of static friction μ be in this case

for the casters not to slip on the surface? (1.0 point)

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Figure 3. The initial conditions of the rod in section 4.

Figure 4. Caster-wheels’ circular movement.

2.2. Solution 

(i) (a) The initial velocity and angular velocity of the rod is

v0 =v1 + v2

2, ω0 =

|v1 + v2| L

.

(b) Since the mass and the radius of the caster-wheels are negligibly small, the force of 

static friction only has a component parallel to the rod (otherwise the component

of static friction perpendicular to the rod will cause an infinitely large angular

acceleration). From this, it follows that the torque acting on the rod is zero, i.e.

the angular velocity of the rod is constant:

ω(t ) = ω0.

Since all the forces acting on the rod are parallel to its axis, the magnitude of the

velocity of the center of the rod remains constant:

|v(t )| = v0,

and the direction of v(t ) is always perpendicular to the rod. So the orbit of the center

of mass will be a circle (see figure 4). During one period (2π /ω), the center of the

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Figure 5. Caster-wheels on a slightly inclined plane.

rod draws a whole circle, so one can write:

2πω0

v0 = 2π R,

where R is the radius of the circle. From this, it follows that the radius of the center

of mass is

 R = v0

ω0

= v1 + v2

|v1 − v2| L

2.

(c) The centripetal acceleration of the center of mass is v20 / R, so the equation of motion

of the rod reads

μmg = mv0ω0,

and the minimal coefficient of static friction for the slipless motion is

μ

=

(v0ω0 )

g=

v2

1 − v22

2gL

.

(ii) (a) Figure 5 shows the forces acting on the rod in the plane of the surface when the angle

between the rod and the steepest line is ϕ. Since there is no torque acting on the rod,

the angular velocity remains constant: ω(t ) = ω0. The equation of motion of the

center of mass in the direction perpendicular to the rod is

mv = −mg sin α sin ϕ.

Taking into account that ϕ = ϕ0 + ω0t  (where ϕ0 is the initial angle between the rod

and the steepest line), the equation of motion is analogous to the differential equation

of a harmonic oscillator. So the magnitude of the velocity of the center of the rod as

a function of time is given by

v(t )=

v(0)+

g sin α

ω0

cos(ω0

t +

ϕ0

),

and its direction is always perpendicular to the rod. The x - and y-component of the

velocity reads

v x (t ) =

v(0) + g sin α

ω0

cos(ω0t + ϕ0 )

cos(ω0t + ϕ0 ), (1)

v y(t ) =

v(0) + g sin α

ω0

cos(ω0t + ϕ0 )

sin(ω0t + ϕ0 ). (2)

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Figure 6. Sketch of the caster-wheels’ center of mass position.

Taking the time average of the components (so we can eliminate the circular motion

of the center of the rod):

v

 x (t ) =

g sin α

ω0 cos

2(ω0

t +

ϕ0

) =

g sin α

2ω0

,

v y(t ) = 0.

Now it is obvious that the drift velocity is

vdrift =g sin α

2ω0

= gL sin α

2|v1 − v2|,

and its direction is horizontal (i.e. perpendicular to the steepest line).

(b) A sketch of the orbit of the center of mass can be seen in figure 6.

(iii) (a) Equations (1) and (2) are still valid, we only have to fit them to the initial conditions:

v(t ) =

v0 +g sin α

ω0

(cos(ω0t ) − 1)

cos(ω0t )

sin(ω0t )

.

(b) The velocity of the center of the rod is zero if 

cos(ω0t ) = 1 − v0ω0

g sin α.

Since−1 cos(ω0t ) 1, the condition of the stopping of the center of the rod reads

v0ω0

g sin α 1.

(c) We have to distinguish the following two cases: the center of the rod stops for a

moment during the motion or it does not. In the first case, the conservation of the

mechanical energy tells us that the maximal displacement along the y direction is

 ymax =v2

0

2g sin α.

In the second case

when v0ω0g sin α > 1

, the center of the rod has a finite velocity even at

the highest point of its orbit. From expansion of v(t ) one can see, that at the highest

point of the orbit the velocity of the center of the rod is

v x 

t = π

ω0

= v0 −2g sin α

ω0

,

so the conservation of mechanical energy reads12

mv20 = mgymax sin α + 1

2mv2

 x ,

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Figure 7. Caster-wheels’ cycloid movement.

from which one get the result

 ymax =2(v0ω0 − g sin α)

ω20

.

(iv) (a) In this situation, using the equations (1) and (2) we can get the velocity of the rod’s

center as the function of time:

v(t ) =

g sin α

ω0

sin(ω0t )

sin(ω0t )

− cos(ω0t )

,

which after some trigonometric manipulations we get:

v(t ) =

g sin α

2ω0

1

0

−

cos(2ω0t )

sin(2ω0t )

.

It can be seen that the endpoint of the velocity vector moves along a circle with radius

v∗ = g sin α

2ω0and center

g sin α

2ω0, 0

. So the center of the rod moves as a point at the

perimeter of a rolling circle with velocity v∗, which is a cycloid. We can calculate the

radius of this ‘rolling circle’, if we recognize that the period of this motion is π /ω0:

r ∗ = 1

2π

π

ω0

v∗ = g sin α

4ω20

.

The difference between the lowest and the highest positions of the center of the rod

is  y = 2r ∗ (measured along the steepest line). The orbit of the center and the two

endpoints of the rod is shown in figure 7.(b) The magnitude of the static friction force is maximal if the center of the rod is at the

lowest possible position during the motion. At this point, the centripetal acceleration

of the center of mass is (v∗)2/r ∗, so the equation of motion is

μmg cos α − mg sin α = m(v∗)2

r ∗,

so μ = 2 tan α.

3. Experimental problem 2: ‘A rotary magnetic drag system for conductivity

measurement’

3.1. Problem statement 

3.1.1. Apparatus (figure 8  ).

(1) Rotary magnet assembly + two polyfoam spacers

(2) Copper plate (1 pc)

(3) ‘Black box’ containing unknown metal X (1 pc)

(4) Thin cardboard spacers (5 pcs)

(5) Thick cardboard spacers (2 pcs)

(6) Power supply unit (1 pc)

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S22 H J Kwee et al 

Figure 8. Left: experimental equipment. Right: the rotating magnet assembly.

(7) Digital multimeter (DMM) generic (2 pcs)

(8) DMM with frequency measurement (1 pc)

(9) Ruler (1 pc).

 Im portant experimental data

Copper plate thickness : t  = 0.6 mm

Copper plate conductivity : σ  = 6.0 × 107 ( m)−1

Metal X thickness : t X = 1.05mm

Thin cardboard thickness : d 1 = 1.10mm

Thick cardboard thickness : d 2 = 2.35mm

Magnet puck location from center of the disc : R = 13mm

3.1.2. Introduction. Electrical conductivity measurement of a material is important for many

applications, such as the metallurgy and semiconductor industries. Usually one measuresconductivity (σ ) (1/resistivity) by a simple resistance measurement that requires making

electrical contacts. This is troublesome and sometimes impossible due to the presence of 

insulating layer. Thus a contactless conductivity measurement technique is desired. In this

problem, we will explore a simple and fascinating system to perform contactless conductivity

measurements utilizing the magnetic drag or braking effect that occurs between fast-moving

magnets and a metal sheet.

If a magnet moves with velocity v parallel to the plane of a non-magnetic conducting

material with conductivity σ  and thickness t , it will experience a magnetic braking effect, also

known as the eddy current braking effect, as shown in figure 9. The magnetic braking force

(using a magnetic dipole model and a thin metal sheet approximation) is given as:

F MB

= −α σ  t 

v

d m

, (3)

where α is the magnetic braking coefficient  of the system which depends on the magnetic

moment of the magnet and the magnetic permeability of the metal sheet (in this experiment

the magnetic permeability of metal sheets is taken to be equal to the magnetic permeability

of vacuum), σ  and t  are the conductivity and thickness of the metal, v is the velocity of the

moving magnet, d  is the distance between the center of the magnet and the metal and m is the

distance power law factor  to be determined in this experiment. The negative sign indicates a

force that opposes the velocity and is thus called ‘magnetic braking’.

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Figure 9. The magnetic braking effect of a moving magnet near a metal sheet.

In this experiment we mount two strong magnetic pucks on a rotating disc driven by a

motor, as shown below. When the disc rotates and the metal plate is inserted underneath, the

disc will slow due to the magnetic braking effect. This effect can be exploited to measure the

conductivity (or thickness) of a metal sheet.

System information

(1) The motor is driven by a variable voltage power supply with coarse and fine control.

(2) DMM#1 measures the motor voltage V  M . DMM#2 measures the motor current I  M  as a

voltage across shunt resistor RS . RS  is different for every set-up. You have to read the label

on the resistor.

(3) The Hall sensor serves as rotational frequency sensor. It provides a voltage pulse each

time a magnet passes by. When the disc rotates, the Hall sensor frequency f  H  can bemeasured using DMM#3.

(4) To demonstrate the magnetic braking effect, a metal plate is placed at a distance d  from

the rotating disc. Note that d  is measured from the center of the magnets (see figure 10).

Thick and thin cardboard spacers are provided to vary the distance d .

(5) There are two metals provided: a copper plate (with a known conductivity σ  and thickness

t ) and an unknown metal X inside a ‘black box’ (with a known thickness t X). See

section 3.1.1: ‘important experimental data’.

 Attention

(1) Do not change the wiring/experimental set-up.

(2) The disc can rotate up to f  H  = 200 Hz. In general, useful data can be obtained

approximately <120 Hz. Do not run the motor at maximum frequency (around 200 Hz)

for too long.(3) To save battery power, do not run the motor unnecessarily. Turn off the power if not in

use.

3.1.3. Experiment and questions. This experiment is divided into four sections.

(1) Introduction: speed sensor and basic operation (1.5 points)

(2) Basic system characterization (2 points)

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S24 H J Kwee et al 

Figure 10. Connection diagram of the rotary magnet set-up.

(3) Magnetic braking effect characterization (4.5 points)

(4) Conductivity measurement of an unknown metal X (2 points)

(1) Introduction: speed sensor and basic operation (1.5 points). This section will help to

familiarize you with the experimental set-up. First we will try to understand the operation

of the Hall effect sensor that measures the rotation frequency. We will observe the voltage

signal coming from the Hall sensor.

(a) Set DMM#3 to dc voltage mode. Turn the rotating disc (mounted on the motor)manually so that the magnet passes the Hall sensor.

Sketch the voltage waveform for two full rotations, mark how the voltage

changes with or without magnet near the sensor. Indicate the period of the

waveform. (0.25 points)

(b) Now switch DMM#3 to frequency mode (Hz) to measure the Hall sensor frequency

 f  H . No metal plate underneath. Turn on the motor power supply and increase the

voltage gradually to speed up the rotation. Operate the motor <150 Hz. Observe how

the frequency reading f  H  increases with speed.

 Express the disc angular frequency ω (in rad s−1) in terms of Hall sensor frequency

 f  H . (0.25 points)

(c) Now insert the copper plate. Note how the rotation slows due to the magnetic braking

effect. How does the magnetic braking effect work? To help answer this question,

consider only interaction between the moving magnet and an ‘elemental ring’ from

the metal sheet, as shown in figure 11.

 Draw all the necessary electromagnet fields involved in this diagram on the answer 

sheet. (1.0 point)

(2) Basic system characterization (2 points)

The motor has an internal series resistance R M  which is not negligible in this experiment .

 R M  is the sum of resistance of the rotor coil inside the motor. Therefore when the motor

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Figure 11. Magnet over an ‘elemental ring’.

Figure 12. The equivalent circuit of the motor.

is driven by a voltage source, not all of the power is converted to kinetic or rotational

energy. A fraction of the energy is turned into heat due to R M . Thus the real motor can be

modeled as an ideal motor (whose coil has no resistance) plus a series resistance R M  as

shown in figure 12.

(a) Determine the internal series resistance of the motor R M  to at least two significant 

 figures. Note. The series resistance is very small. Direct resistance measurement

using the ohmmeter mode of the DMMs does not have enough accuracy. Try another

method. You do not have to change the experimental set-up. Please also write down

the value of shunt resistor RS  (the resistor to measure motor current I  M ) on your

answer sheet. (0.7 points)

Now we will explore how the system behaves at various ranges of rotation speeds with no

metal plate inserted . This is important to pick the operating range for further experiment.Under normalcondition the motor should runsmoothly, however around some frequencies

the system vibrates strongly and becomes noisy and we want to avoid this. Do not insert 

the copper plate.

(b) Increase the motor voltage gradually such that the frequency rises from 0 to near

200 Hz, record the Hall sensor frequency f  H  and the current I  M  (DMM#2). Plot both

data with respect to voltage V  M . Mark the range of frequencies where the system

vibrates strongly and become noisy. (0.8 points)

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S26 H J Kwee et al 

Figure 13. The cross section of the ‘black box’ containing the unknown metal X.

(c) What might cause this strong vibration/noise at certain frequencies? What anomaly

do you observe in the motor current I  M  versus voltage V  M  plot in this noisy

region? (0.5 points)

(3) Magnetic braking effect characterization (4.5 points)

We now study the characteristics of the magnetic braking effect with the copper plate

provided. With the knowledge from the previous section, we want to operate the system

in the quiet region as much as possible, not in the noisy region.

(a) Derive the expression for power dissipation PMB. Due to the magnetic braking force

of the two magnets and the metal plate. Use equation (3) and your answer from

question (1)b. Express PMB in terms of the Hall sensor frequency f  H . (0.5 points)

(b) Perform experiments to verify the relationship of PMB and f  H  from your previous

answer. (1.5 points)

(c) Perform experiment to determine the magnetic braking force coefficient α (note that

α is associated with one magnet) and distance power law factor m. (2.5 points)

 Note. to maintain the validity of the distance power law in equation ( 3) (due to

magnetic dipole approximation) do not place the metal too close. Maintain sufficient

distance, approximately d  > 8mm.

(4) Conductivity measurement of an unknown metal X (2 points)

To obtain conductivity (assuming the thickness is known), the distance between the magnet

and the metal plate d  needs to be determined accurately because the magnetic braking

force falls off strongly with distance due to a large power factor m. In actual industrialapplications, it is difficult or even impossible to obtain an accurate distance d  without

making contact with the metal (remember this is a contactless method).

However, it is possible to perform the conductivity measurement without knowing

the exact distance between the magnet and the metal plate d . This problem simulates this

situation. We have an unknown metal X inside a ‘black box’, as shown in figure 13. The

thickness is known:

t X = 1.05mm.

However, its exact location from the surface (d 0 ) is not known. In fact, you do not need

to know it to obtain the conductivity σ . You may assume that α in this part is the same as

part (3)c.

(a) Perform an experiment to determine the conductivity of the metal

 X (σ  X ). (2.0 points)

 Note

Please do not open the ‘black box’. It will invalidate your answer.

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Figure 14. Voltage waveform.

Figure 15. Magnet over an ‘elemental ring’.

3.2. Solution 

1. System introduction: speed sensor and basic operation (1.5 points)

1. (a) Turn the rotating disc (mounted on the motor) manually so that the magnet passes

through the Hall sensor. Sketch the voltage waveform for two full rotations, mark 

how the voltage changes with or without magnet nearby sensor. Indicate the period

of the waveform. (0.25 points)

Answer: see figure 14.

1. (b) Express the disc angular frequency ω (in rad s−1) in terms of Hall sensor frequency

 f  H  (0.25 points)

ωDISC = 2π f Hall

2= π f Hall

The factor of 1/2 multiplying f Hall because there are two magnets giving two pulsesper rotation. Thus the disc rotation frequency is half of the Hall sensor frequency.

1. (c) How does the magnetic braking effect work? To help answer this question, consider

only interaction between the moving magnet and an ‘elemental ring’ from the metal

sheet, as shown below. Draw all the necessary electromagnet fields involved in this

diagram on the answer sheet. (1.0 point)

See figure 15. Consider a segment of the ring P right underneath the magnetic puck 

where the magnetic field is the strongest.

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S28 H J Kwee et al 

Figure 16. Determination of motor internal series resistance R M .

1. The magnet moves, creating an increasing magnetic flux in the ring, thus inducing

an electromotive force (EMF) voltage V EMF according to Faraday’s law.

2. The EMF voltage produces current I  in the ring whose direction is determined

by Lenz’s law such that the induced magnetic field ( Bind) opposes the change of 

the magnetic field from the magnet.

3. This current flow I  and the magnetic field B from the magnet produces a Lorentz

force F  R on the ring segment under the magnet with the same direction as the

velocity of the magnet.

4. Due to Newton’s third law, the Lorentz force F  R is accompanied by a reactionforce F MB acting on the magnet in the opposite direction. This is the magnetic

braking force.

2. Basic system characterization (2.0 points)

2. (a) Determine the internal series resistance of the motor R M  to at least two significant

figures. (0.7 points)

The shunt resistance for motor current measurement is RS  = 1.11. Note that this

value may vary approximately±20% for different set-ups. The motor current is given

as I  M  = V 2/ RS , where V 2 is the DMM#2 voltage.

For R M  measurement the motor should not rotate, so that there is no voltage drop due

to back EMF from the rotor coil due to rotation (V EMF = − BAω N ).

V  M  = V EMF + I  M  R M 

Thus if we apply the voltage across the motor, the entire voltage drop occurs across

the internal series resistance R M . This can be achieved by applying very small voltage

(e.g. V  < 100 mV) so the motor will not rotate, or by holding the motor.

 R M  is very small (∼0.8), thus we need to plot V  M  versus I  M  (or V 2/ RS ) to obtain

 R M , see figure 16.

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Figure 17. Motor resonance region.

We note that the linearity of the data is good, and yield:

 R M  =V  M 

 I  M 

= V  M 

V 2 RS  = 0.855 .

2. (b) Increase the motor voltage gradually from 0 to 1.7 V, record the Hall sensor frequency

and the current (DMM#2). Plot both data with respect to voltage. Mark the range of 

frequencies where the system vibrates strongly and become noisy. (0.8 points)

Both plots are shown in figure 17.

2. (c) What may cause this strong vibration/noise at certain frequencies? What anomaly

that you observe in the motor current ( I ) versus voltage (V ) plot corresponds to thisnoisy region? (0.5 points)

This strong vibration or noise happens because of the resonance effect, which is

around ∼120–180 Hz in this set-up. At resonance there is maximum energy transfer

or efficient mechanical coupling from the motor to the chassis.

This resonance effect is apparent in the ‘bump’ of the current versus voltage plot

around the resonance frequency, which indicates larger power dissipation (V  M × I  M )

near the resonance.

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S30 H J Kwee et al 

Figure 18. Power dissipation.

 Note. This resonance effect has some variations between set-ups, nevertheless it

occurs mostly at ∼100 Hz and above. There is actually also a hysteresis effect

present. But we shall ignore this issue for the moment.

3. Magnetic drag effect characterization (4.5 points)

3. (a) Derive the expression for power dissipation PMB due to the magnetic braking force

of the two magnets and the metal plate. Use equation (3) and your answer from

question 1(b). Express PMB in terms of the Hall sensor frequency f  H . (0.5 points)

The power dissipation due to the magnetic braking force of the two magnet pucks is

PMB

= −2F MBv

=2ασ t v2

d m,

where the factor of two is due to two magnetic pucks and the negative sign is due to

the force opposite to the velocity, in other words this is a dissipative force. The radial

velocity of each magnet, using the answer from A.2, is: v = ω R = π f  H  R.

Thus finally:

PMB = 2ασ t π 2 f 2 H  R2

d m.

We omit the negative sign − with the understanding that this is a dissipative force.

Note that the full expression for the magnetic drag or braking force can be obtained

from [1] at the low velocity limit ν 2/μ0σ t , which yields:

α = 3μ20m2

 M 

128π

and m = 4, where μ0 is the magnetic permeability and m M  is the magnetic momentof the magnet.

3. (b) Perform experiments to verify the relationship of  PMB with f  H  from your previous

answer. (1.5 points)

We recognize that the motor slows if the copper plate is inserted underneath, in other

words it takes more power for the system to rotate at the same frequency, see figure 18.

Thus the power difference for the system to operate at the same frequency is due to

the magnetic braking effect.

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Figure 19. Magnetic power dissipation as a function of frequency.

However, we should recognize that there is also power loss due to the motor’s internal

series resistance R M  and this has to be accounted for.

We can write: PMB = P = PK − PK , where P

K  and PK  are the motor kinetic power

with and without the metal being inserted. This motor kinetic power is net power

delivered to the motor after discounting the Joule loss due to R M , and can be calculated

as:

PK  = V  M  I  M − I 2 M  R M .

Therefore the power dissipation PMB can be obtained from measurement of V  M , I  M 

(with metal), and V  M  and I  M  (without metal) while maintaining a fixed frequency:

PMB = V  M  I  M − I 2 M  R M − V  M  I  M − I 2 M  R M = 1

 RS 

(V  M V 2 −V  M V 2 ) −

V 22 −V 22

 R M 

 RS 

,

because I  M  = V 2/ RS .

Note that the student might perform the experiment with a series of fixed voltage

settings and record the current and Hall sensor frequency with and without the

copper plate. However, an astute student will recognize that analysis is easier if the

measurements are performed at fixed frequency settings.

This experiment can be performed as follows.

1. Without the metal plate, choose a fixed distance d and choose a rotation frequency

 f  H . Record V  M  and I  M .

2. Insert a metal plate and the system will slow down.

3. Increase the voltage or speed accordingly to match the original frequency.

4. Record V  M  and I  M .5. Repeat steps 2 to 4 at various frequencies outside resonance.

The plot is shown in figure 19. The plot of √ 

PMB versus f  H  is linear, indicating that

PMB ∝ f  H , consistent with the result in (1)a.

3. (c) Perform the experiment to determine the magnetic braking force coefficient α (note

that α is associated with one magnet) and distance power law factor m. (2.5 points)

In this experiment, we have to vary the distance d  to obtain both α and m. This

experiment can be performed as follows.

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S32 H J Kwee et al 

Table 1. Distance dependence of the magnetic braking power.

d V  M  I  M  PK  PMB

(mm) (V) (A) (mW) (W) ln(d ) ln(PMB) Note

13.00 0.353 0.1009 2.596 – – – No metal

13.00 0.366 0.1153 3.391 0.7949 −4.343 7.137 With metal11.95 0.373 0.1225 3.828 1.2320 −4.427 6.699 Metal + 1 thin10.90 0.380 0.1306 4.351 1.7554 −4.519 6.345 Metal + 2 thin10.65 0.387 0.1324 4.472 1.8763 −4.542 6.278 Metal + 1 thick  

9.85 0.394 0.1387 4.908 2.3124 −4.620 6.069 Metal + 3 thin9.60 0.401 0.1486 5.635 3.0386 −4.646 5.796 Metal + 1 thick + 1 thin8.80 0.408 0.1568 6.266 3.6700 −4.733 5.608 Metal + 4 thin8.55 0.425 0.1685 7.237 4.6413 −4.762 5.373 Metal + 1 thick + 2 thin8.30 0.431 0.1748 7.789 5.1933 −4.791 5.260 Metal + 2 thick  

1. Choose a starting distance d  with no spacers. Choose an operating frequency,

medium speed frequency ∼40 Hz is good (not too low and not too high).

2. With no metal inserted, record V  M  and I  M .3. Insert the metal plate, increase the voltage to match the original frequency. Record

V  M  and I  M .

4. Insert spacers, record the new distance d , repeat step 3.

Note that the distance must be kept above approximately 8 mm, to maintain the

validity of equation (3) due to the magnetic dipole model.

We pick a medium frequency of 60 Hz to allow more data points (if we choose too

high, the power supply may not have enough drive at higher load (when the metal is

very close). Frequency: f  H  = 60 Hz.

Note that PMB = PK  − PK , where P

K  and PK  is the motor kinetic power after and

before the metal is inserted. PK  is given as:

PK 

=V  M  I  M 

− I 2 M  R M .

To extract α and m we use the following linear regression ( y =  A + Bx ) analysis:

PMB = 2ασ t π 2 f 2 H  R2

d m,

ln PMB = ln

2ασ t π 2 f 2 H  R2− m ln d .

The data can be found in table 1 and the plot is shown in figure 20 (in SI units).

Finally we obtain:

α = e A

2σ t π 2 f 2 H 

 R2

= e−24.54

2 × 6 × 107 × 0.6 × 10−3 × π 2 × 0.0132 × 602

=5.09

×10−17 Nm3s,

m =  B = 4.0.

Note that in this experiment we obtain the power distance exponent factor m = 4.0

in excellent agreement with the theory [1]. Repeated experiments put the factor near

this predicted value.

4. Conductivity measurement of an unknown metal X (2 points)

4. (a) Perform an experiment to determine the conductivity of the metal X. (2 points)

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Figure 20. Log of magnetic power braking versus log of distance.

Table 2. Magnetic power braking to the power of −1/m versus distance.

d V  M  I  M  PK  PMB

(mm) (V) (A) (mW) (W) P−1/m

MB Note

13.0 0.417 0.1063 2.882 – – No black box (BB)13.0 0.400 0.1234 3.885 1.0025 5.6199 With BB11.9 0.412 0.1297 4.292 1.4096 5.1609 BB + 1 thin10.8 0.410 0.1360 4.719 1.8370 4.8303 BB + 2 thin

9.7 0.450 0.1505 5.772 2.8900 4.3130 BB + 3 thin8.6 0.458 0.1649 6.931 4.0490 3.9643 BB + 4 thin7.5 0.465 0.1937 9.567 6.6849 3.4972 BB + 5 thin

The key idea in this experiment is that we can still extract the coefficient α without

knowledge of the absolute d  value, i.e. by varying or stepping the apparent distanced . Both d  and d  are related as:

d = d  + d 0.

This can be done by recasting the PMB equation into:

P−1/mMB =

2ασ t π 2 f 2 H  R

2−1/m

(d  + d 0 ).

The value β (and thus σ X) can then be obtained by linear regression of  P−1/mMB versus

d . We perform a similar measurement as in the previous section, maintaining a fixed

rotation frequency ( f  H  = 60 Hz). Since the metal X is at unknown distance d 0within the black box, we can use an arbitrary reference to record the distance d . A

sample of data can be found in table 2. From linear regression we obtain A = 0.6452

[SI] and B = 382.4 [SI] (see figure 21). Finally with B =

2ασ Xt π 2 f 2 H  R2−1/m

, we

obtain:

σ X = 2α Bmt π 2 f 2 H  R2

−1

= 7.3 × 107(m)−1.

During WoPhO final round, we sampled four stations and arrived at the following

range of accepted values:

σ X = (0.3 − 7) × 107 (m)−1.

Most of the error is due to inaccuracies caused by the spacer not being flat. Note that

the intercept at x -axis gives d 0, but this is not raised in the problem.

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S34 H J Kwee et al 

Figure 21. Magnetic power braking to the power of −1/m versus distance.

4. First WoPhO final round results and discussion

The first WoPhO final round awarded 11 gold medals, 12 silver medals, and 11 bronze medals.

Complete results are available online (see footnote 5). Unlike other olympiads, no honorable

mentions were made. The highest overall score was 35.10 from a possible score of 50.0. This

score was achieved by Kexin Yi, the overall winner of the first WoPhO. He also achieved

the highest score for the theoretical examination, receiving 22.75 points from a possible 30.0.

The highest score for the experimental examination was 13.50 from a possible 20.0, achieved

by Eugen Hruska. The average score for all the medalists was 17.32, with a theoretical score

average of 9.59 and an experimental score average of 7.73.

We have two comments.

1. The first WoPhO might be considered to be an olympiad with more difficult problems

than previous APhO and IPhO competitions, judging from the scores achieved by the

medalists.

2. Almost all of the medalists in this WoPhO were medalists in previous olympiads

(APhO/IPhO).

References

[1] Reitz et al 1970 J. Appl. Phys. 41 2067[2] Kalda J 2013 Eur. J. Phys. 34 S3–14[3] Kalda J, Kikas J, Heidelberg M, Ainsaar S and Lohmus R 2013 Eur. J. Phys. 34 S35–48