physics competitions vol. 14 no 2 2012 the final...

Physics Competitions Vol. 14 No 2 2012

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The Final Round of the First World Physics Olympiad

held in Lombok, West Nusa Tenggara Indonesia: Problems and Results

Herry J. Kwee and Yohanes Surya Faculty of Physics Education Department, Surya College of Education, Banten, Indonesia

Abstract A brief report of the final round of the first World Physics Olympiad (WoPhO) held in Lombok, West Nusa Tenggara, Indonesia is presented. The theoretical and experimental problems are presented and the mark distribution is discussed.

— 1. Introduction

The final round of the first WoPhO was held in Lombok, West Nusa Tenggara, Indo-nesia from December 28th, 2011 to January 3rd, 2012 and was organized by Surya College of Education and 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 round. The selection round is meant to provide as many opportunities as possible to students to participate in an Olympiad-level physics competition. Problems are presented online and any student who is eligible1 can participate. For the first WoPhO selection round, 345 students from all over the world register to participate. The discussion round provides an op-portunity for the participants and the general public to discuss the problems and solu-tions of the selection round and other physics Olympiad-level problems. And the final round, which is organized very similar to the International and Asian Physics Olympi-ad (IPhO and APhO) provides an opportunity for every participants to challenge the APhO and IPhO gold medalists. Just like in APhO and IPhO, there are 5 hours at the students´ disposal to solve the three theoretical problems and another 5 hours for the experimental tasks. There were 122 students from 13 countries who participated in the first WoPhO final round, even though the distribution was not evenly spread. The majority of students were from the host country, Indonesia. 1 Students who have not started college or reached 20 years of age by June 30th of competition year.


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Another unique characteristic of WoPhO is the fact that the problems for the final round are not provided by the host country but are selected from a competition (or simply called WoPhO problem competition). Anyone can participate in this competi-tion, in fact one of the winners of this first WoPhO problem competition was a high school student and another one was a university student. Three theoretical problems were chosen from thirty one 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. In section 2, we present the theoretical problems. In section 3, we present the experimental problems. In section 4, we present the summary of the result of the final round.

— 2. Theoretical Examination

Theoretical Problem No. 1: Motion of a Rolling Rod

In this problem, the motion of a uniform rod (stick) with length L , ended 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 diameters of the caster-wheels are a bit larger than the diameter of the rod, but both diameters are much smaller than the length of the rod. The gravitational accel-eration is g .

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

The rod is placed on a horizontal flat surface and pushed such that each end of the rod get different horizontal initial velocity ( 1v and 2v , pointing in the same direction)

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


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Calculate the initial velocity 0v of the center of the rod and the initial angular ve-

locity 0ω of the rod using 1v , 2v and L ! [0.8 points]

Describe the motion of the center of mass of the rod! Determine the parameter(s) of its orbit! [0.8 points]

What should be the minimum value of the coefficient of static friction µ for the

casters to not slip on the surface? [0.6 points]

In the following sections the case of the inclined surface will be considered. The an-gle between the inclined surface and the horizontal plane is α . 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 a constant drift velocity driftv added to the solution. Use a coordinate system as in Figure 2.

Calculate the magnitude and the direction of driftv as a function of the small α ,

the initial velocities of the two ends of the rod ( 1v and 2v , pointing in the same di-

rection) and the gravitational acceleration g ! [1.9 points]

Sketch the orbit of the center of mass of the rod! [0.5 points]

If α is finite, the details of the motion of the rod changes. Place the rod on the in-clined plane along the steepest line of the surface (so the rod is parallel with the in-clined edges of the plane). Consider that the initial velocity 0v of the center of mass of

the rod is perpendicular to the axis of the rod and the initial angular velocity 0ω is

perpendicular to the surface as shown in Figure 2.

Figure 2: The initial conditions of the rod

Calculate the time evolution of the ve-locity ( ) ( ( ), ( ))x yt vt v t=v of the center of mass

of the rod in the Cartesian coordinate system shown in Figure 2. [0.8 points] Depending on the magnitude of the 0v

and 0ω , it can occur, that the center of the rod

stops for a moment during its motion. Express the condition(s) for such a behavior using the parameters 0v , 0ω , g , α and L ! [0.8 points]

Determine the maximum displacement of the center of the rod in the direction of the steepest line ( y -direction) as function of 0v and 0ω ! [1.2 points]


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Investigate another situation where the rod is placed horizontally on the inclined sur-face. Consider that the initial angular velocity 0ω of the rod is perpendicular to the

surface but the initial velocity of the center of the rod is zero (see Figure 3).

Figure 3: The initial conditions of the rod

Describe the motion of the center of mass of the rod! Determine the pa-rameter(s) of its orbit! [1.6 points] What should be the minimum val-ue of the coefficient of static friction µ in

this case for the casters to not slip on the surface? [1.0 points]

— Theoretical Problem No. 2: Why Maglev Trains Levitate

Maglev is a technology for magnetic suspension (levitation) and propulsion of trains or other vehicles. Since there is no friction force between the train and the rails, Mag-lev trains reach record velocities approaching 600 /km h .

There are three types of Maglev systems -- EMS (Electromagnetic Suspension), EDS (Electro-dynamic Suspension), and the experimental Inductrack technology. In this problem you are going to explore the physical principles of Inductrack suspension on a simplified model of a Maglev train. The principle of magnetic propulsion will not be considered here because it has a lot in common with the physics of magnetic levita-tion. Shown in Figure 4 is a schematic side view of an Inductrack train-car. When at rest or moving at a low speed, the car lies with its wheels on the rails like any ordi-nary train. The car, however, detaches from the rails at a specific takeoff velocity 𝑣𝑡 due to the system described below: two long parallel arrays of permanent cubic-shape magnets of size a = 5 cm, each is located at the bottom of the car. The mag-netic dipole moments of the neighboring magnets are tilted at an angle of 45° relative to each other (the so called Halbach array). As a result a static magnetic field is pro-duced below each array with components of the magnetic field given by the equa-tions:

0 exp( )sin( )xB B ky kx= − (2.1)

0 exp( )cos( )yB B ky kx= − (2.2)


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where the x-coordinate is measured from the rear end of the car in the direction of motion and y-coordinate from the bottom of the car in a downward direction. The pa-rameter k is the wavenumber of the magnetic field. The amplitude of the magnetic field is 0 1.4 B T= .

Figure 4: A schematic side view of an Inductrack train-car. The arrows show the directions of magnetic moments of the cubic magnets in the Halbach array. The wheels and the rails of the train are not shown for convenience.

Two inductive arrays of horizontal rectangular wire frames of the same width a made of permanent magnets are arranged along the rail as seen from the top view Figure 5. The length of the inductive frames as well as the distance between them is b. Each inductive array is located below the corresponding Halbach array at a dis-tance corresponding to the y-coordinate of the inductive array (see Figure 4). As-sume that the equations (2.1) and (2.2) for the components of the magnetic field hold only for the inductive frames, which are located below the car, while the magnetic field outside the car area is strictly zero.

Figure 5: A schematic top-view of an Inductrack train-car. The wheels and the rails of the train are not shown for convenience.

Note: In the real Inductrack trains, the frames of the inductive array are electrically connected and form a continuous ladder-like array. In order to simplify the theoretical consideration, we use a simplified model shown in Figure 5.

Helpful mathematics formulas:

sin( ) sin( ) 2sin( ) cos( )2 2

x y x yx y − +− = (2.3)


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cos( ) cos( ) 2sin( )sin( )2 2

y x x yx y − +− = (2.4)

Assume that the train car is infinitely long (i.e. equations (2.1) and (2.2) hold for all values of the x-coordinate.

a) Electromotive force in the inductive wires Derive a relationship between the wavenumber of the magnetic field k and the

lateral size of the cubic magnets in the Halbach array a. Calculate k numerically. [0.3 points]

Suppose that the train moves with a constant velocity v in the positive direction of the x-axis. Consider an inductive frame whose center in the initial moment 0t = is located below the car at a point of coordinate cx relative to the car. Derive an

expression for the electromotive force (EMF) ε induced in the frame as a func-tion of the time t and the other parameters already defined. [1.5 points]

What is the angular frequency ω of the induced EMF. [0.4 points]

b) Current in the inductive wires Each frame in the inductive array is characterized by a self-inductance L and a re-sistance R. The mutual inductance between different frames is negligible compared to L. The current I induced in the frame varies with time according to the equation: 0( ) sin( )cI t I t kxω ψ= − +

Obtain expressions for the amplitude 0I and the phase-shift ψ of the alternat-

ing current induced in the frame. [2.0 points] Note: The positive direction of circulation of the induced current in the frame is related to the positive direction of the y-axis according to the right-hand rule.

c) The dynamics of the train Derive formulas for the time-averaged components xF and yF acting on a single

inductive frame in terms of velocity v, the distance between the Halbach array and the inductive array y, and the others parameters already defined. Sketch qualitative graphs of xF and yF versus train velocity v for a fixed value of the dis-

tance y. [3.0 points] Consider a train-car of a large but finite length (l and b). Derive expressions for

the magnitude of the total lift (vertical) and drag (horizontal) forces LF and DF

acting on the car. [1.2 points]


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For a given value of a, what is the minimum aspect ratio /b a for which the lift force is maximum for given values of the distance y, velocity v, resistance R, and the inductance L. [1.0 points]

Consider a train-car of length 10 l m= and mass 10000 kgm = . Assume that the

aspect ratio /b a corresponds to the optimal value found in 3(c). The inductance

and the resistance of the inductive frames are 71.0 10 L H−= × and 51.0 10 R −= × Ω

respectively. The acceleration due to gravity is 29.8 /g m s= . Assume that the dis-

tance between the Halbach array and the inductive array is 0 y m= , when the

train is at rest. Obtain an expression and calculate the takeoff velocity tv , the velocity at which

the train detaches from the rails. [0.3 points] What is the distance y between the train and the rails at an operating velocity of

360 km/h?

— Theoretical Problem No. 3: Mirage

The refractive index of the air varies with temperature. Cold air is denser than warm air and has therefore a greater refractive index. Thus a temperature gradient in the atmosphere is always associated with a gradient of the refractive index. Under cer-tain conditions, this gradient of the refractive index could be strong enough so that light rays are bent to produce a displaced image of distant objects. This amazing phenomenon is called a mirage. Mirages can be categorized as "inferior" and "superior". Inferior mirages can be seen on deserts and highways, and superior mirages occur over the sea. To describe in detail the phenomenon of the mirage, we need to analyze the path of light rays in media with a refractive index gradient.

Figure 6: Left: A superior image. Right: An inferior image


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The path of a light ray and the trajectory of a mass point Media with a refractive index gradient can be treated as being formed from thin ho-mogeneous layers with different refractive indices. The path of a light ray can be de-termined then by analyzing refractions of the light ray at interfaces between these thin homogeneous layers. But there exists a more convenient way for determining the path of a light ray in media with a refractive index gradient. Instead of analyzing the propagation of the light, one may study the motion of a mass point that moves along the path of the light ray, driven by a conservative force. The potential energy of the conservative force field depends on the distribution of the refractive index. Once the potential energy is established, one may study the motion of the mass point by using well developed tools in classical mechanics, and find the trajectory of a mass point which is also the path of the light ray.

Refraction of a light ray at the interface between a medium with a refractive index 1n

and a medium with a refractive index 2n is shown in Figure 7a. The angles 1i and 2i

obey the Snell's law: 1 1 2 2sin sinn i n i= . (2.1)

The path of a mass point in a conservative force field is illustrated in Figure 7b. The speed of the mass point is 1v in the region with the potential energy 1pE , and 2v in the

region with the potential energy 2pE .

Find expressions for 1v and 2v so that the relation 1 1 2 2sin sinn i n i= holds also for the trajectory of the mass point. You may use an arbitrary constant expression

0v with the dimension of speed in your expressions. [0.2 points]

Figure 7:

(a) Left: Refraction of a light ray. (b) Right: Deviation of a mass point in a conservative force field


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Assume that the mass of the mass point is m, and the total energy of the mass point equals to zero. Find the potential energies 1pE and 2pE in 0v , 1n and 2n

[0.2 points] The trajectory of a mass point with a mass m in a conservative force field is the

same as the path of a light ray in a medium with a refractive index ( )n r which is

a function of the position. The total energy of this mass point is zero. Find ex-pressions for the potential energy ( )pE r of the conservative force field and the

speed of the mass point ( )v r [0.4 points]

To describe the motion of a mass point, one expresses the position of the mass point as a function of time: ( )r t . To describe the trajectory of a mass point, one needs to express the position of the mass point as a function of the distance s traveled by the mass point from the start point of the trajectory to the current po-sition: ( )r s . Derive a differential equation for the trajectory ( )r s of a mass point with a mass m and a total energy E that moves in a conservative force field. [0.6 points]

Hint: in a conservative force field, ( )r t satisfies Newtons´ second law:

2

2 ,pd rm Edt

= −∇

(2.2) where ,( )

ˆ ˆ ˆ .

ds v rdt

x y zx y z

=

∂ ∂ ∂∇ = + +

∂ ∂ ∂

(2.3)

Derive the light ray equation (a differential equation for the path of a light ray) by using the results obtained in (c) and (d). [0.6 points]

Hint: 2 ( ) 2 ( ) ( )f r f r f r∇ = ∇

The inferior mirage When an inferior mirage appears, an image of a distant object can be seen under the real object. A direct image of that object is seen because some of the light rays enter the eye in a straight line without being refracted. The double images seem to be that of the object and its upside-down reflection in water. For exhausted travelers in the desert it seems like that there is a lake of water in front of them.

An inferior mirage occurs when a strong positive gradient of refractive index is pre-sent near the ground. We use the following model to describe the variation of the refractive index of the air with elevation:


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2

2 020

for 0( )

for ,n z h z

n zn h z h

αα

+ ≥ ≥=

+ > (2.4)

with 20 1 1n − and 1hα .

Figure 8: The geometry for analyzing an inferior mirage

Find the path of a light ray (i.e. z as a function of x) that enters in an observer's eye at an angle θ (see Figure 8). The height of the observer eye is H. [1.9 points]

Due to the inferior mirage, the observer can see an inverted image of the upper part of a camel at a large distance. The parameters for the refractive index of the air are 5 13.0 10 m α − −= × , 0.50 mh = . The height of the observers eye is

1.5 mH = , and the height of the camel is 2.2 ml = . Find the minimum values for the distance mD between the observer and the camel so that the observer still

can see the inverted image of the upper part of the camel. You may use the ap-proximation 2

0 1n ≈ . [0.7 points]

As a result of the light refraction in the region with refractive index gradient, the observer cannot see the lower part of the camels legs. Find the height of the low-est point ( ml ) on the camel at the distance mD that can still be seen by the ob-

server. [0.7 points] Find the distance d between the observer and the imaginary lake of water.

[0.3 points]


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The superior mirage When a superior mirage occurs, light rays that were originally directed above the line of sight of the observer will reach the observer's eyes. Thus, an object ordinarily be-low the horizon will be apparently above the horizon. A superior mirage occurs when a negative gradient of refractive index is present over a body of water or over large sheets of ice. We use the following model to describe the refractive index of the atmosphere (see Figure 9):

[ ]

[ ]20 0 02

20 0

1 ( ) for ( ) 0( )

1 for ,

( )n r r b r r

n rn b r r b

ββ

− − ≥ − ≥=

− − > (2.5)

where 0r is the radius of the earth, 0b r and 1bβ .

Figure 9: The geometry for analyzing a superior mirage

Find the path of a light ray (i.e. r as a function of φ ) within the range

0( ) 0b r r≥ − ≥ . Use γ , the angle between the light ray and the vertical direction at

the sea level as the parameter of the path. [1.4 points] Hint: The trajectory of a mass point with mass m, angular momentum L and total energy E in a conservative force field with

0pmAE Er

= − + (2.6) is ,1 cos

r ρε φ

=−

(2.7)

Where 22

02 3 2

2( ), and 1 .E E LLm A m A

ρ ε −= = + (2.8)

We also have


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00 02

0 0

1 1 if | | .r r r r rr r r

−≈ − − (2.9)

Find the minimum value of ( )mβ β at which the superior mirage occurs. Use the

following values for 0n and 0r : 0 1n ≈ , 60 6.4 m10r = × . [0.8 points]

Under a certain atmospheric condition, with 100 mb = and 7 16.0 10 m β − −= × , cal-

culate the largest distance MD at which the surface of the sea can be seen by an

observer at an altitude 10 my = (y is the altitude of the observer's eye).

[1.4 points]

Useful formula: 21cos 12

φ φ≈ − for 1φ .

For comparison, calculate the largest distance MD′ at which the surface of the

sea can be seen by an observer at the same altitude 10 my = , when the refrac-

tive index of the air is constant. [0.3 points] Calculate the angular difference ϑ∆ between the apparent horizon when a supe-

rior mirage occurs as and the apparent horizon in a normal day, seen at the same altitude 10 my = . [0.5 points]

— 3. Experimental Examination Experimental Problem No.1: Granular Material

The vast majority of mechanics experiments normally conducted in high school phys-ics laboratories are those that have to do with solid objects such as spheres, springs, carts, and rods. Moreover, these solid objects are additionally assumed to be ideal rigid objects such as the ones usually dealt with in any classroom discussion involv-ing dynamics. This approach unfortunately omits the exploration of another class of solid objects that are rather ubiquitous in real life, granular materials. These are ma-terials consisting of many small grains or particles; while each grain or particle is a solid object, the collective behavior of any group of those grains or particles is usually nothing like their more rigid counterparts. The following experiments depart from the common high school mechanics experiments in that they are designed to explore many aspects of the behaviors of granular materials.

The experiments consist of four independent parts. In the first part, we will investigate the static and dynamic properties of a granular material (dry sand) by measuring the


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angle of repose and the maximum angle of stability. This experiment accordingly gives us the information about coefficient of static friction of the sand grains. The mass flow rate of the dry sand through an orifice is investigated in the second exper-iment. In the third part, we will observe how the sand settles in a fluid. Having ob-tained the settling velocity of the sand sediment, we estimate the average size of a sand particle. Lastly, we model an asteroid impact crater using craters formed by balls dropped into dry, non-cohesive, granular media.

The following physical quantities are given:

Gravitational acceleration at surface of Earth: g = 9.81 m/s2

Average density of Earth's continental crust: ρE = 2.7 g/cm3 Average radius of Earth: RE = 6400 km Density of iron at room temperature: ρiron = 7.874 g/cm3 The explosive power of 1000 tons of TNT: 1 kiloton = 4.184x109 J Viscosity of water at room temperature: η = 8.94x10-4 Pa.s

Table 1: List of available apparatus and materials for experiment

Name Quantity Name Quantity Inclined plane with acrylic box + stand

1 Graduated Cylinder 1

Sand in the container with pink cover 1 Colored (red/blue) sand 1 Containers with different orifice size 6 A glass of water 1 Empty container without orifice 2 Ruler 1 Vernier caliper 1 Flashlight 1 Digital weight scale 1 Graph paper 1 Digital stopwatch 1 Tripod 1 Metal stand with release mechanism 1 Glass ball 1 Pile of sand in plastic basin 1 Spoon 1

ATTENTION: Do not mix different types of sands (sand in the container with pink cover, colored sand, and pile of sand in plastic basin are different.) Use the correct sand for each experiment.

Angle of Repose The hourglass is a fascinating and antique timekeeping device. Its distinctive geomet-rical shape makes it immediately recognizable, and, as some people claim, it has a certain elegance which modern timekeeping devices lack. Nevertheless, the geomet-ric shape of the common hourglass must exist for a reason, since one does not find


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cubic hourglasses. The reason, as it turns out, is that granular materials have a cer-tain parameter called the angle of repose, rθ which is a measure of the final angle

formed by the surface of a pile of granular material when it is poured onto a horizon-tal surface. The angle of repose rθ is measured with respect to horizontal surface.

In this problem, we will investigate the angle of repose of dry sand in a simple exper-iment. Initially, a flat rectangular container is filled with sand and the top surface of the sand is flattened out horizontally. The expanse of sand can be stable when its container is tilted as long as its slope is less than the angle of maximal stability, mθ . As the slope

is increased above mθ , some of the sand grains begin to flow and an avalanche oc-

curs.

Note: Use Sand in the container with pink cover

Figure 10: Configuration to determine maximum angle of stability.

Determine the angle of maximal stability, mθ . [1.0 points]

The angle you have just calculated, however, is not the angle of repose. If the sand is too compact it can affect your measurement result.

You may perform another experiment to determine the angle of repose of the sand, rθ . Draw a schematic diagram of your experiment. [1.0 points]

Determine the intergranular friction coefficient, sµ , which is the coefficient of stat-

ic friction between a sand grain and other sand grains. [0.5 points]


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Flow Rate of Granular Matter The mass flow rate of granular materials through an orifice due to gravitational force is generally independent on the geometry of the container. This statement holds as long as the dimensions of the container are sufficiently large compared to the size of the grains and the diameter of the outlet orifice. Note that it was this fact which, though perhaps unknown to the people of that time, enabled hourglasses to be rela-tively good timekeeping devices. In this case we will determine parameters that can affect the flow rate of sand through a circular orifice. Note: Use Sand in the container with pink cover Let us suppose that the mass flow rate of sand, W, through a circular orifice depends on the density of sand ρ , the acceleration due to gravity g, the orifice diameter D,

and the intergranular friction coefficient sµ . Based on this assumption, the flow rate

fulfills the equation ) ,( thraW C gDµ ρ= (2.1)

where ( )C µ is a dimensionless empirical factor which depends on the internal friction

coefficient of the sand and thra is a constant.

Find the theoretical value of thra . [0.5 points]

Measure the flow rate of sand using various orifice diameters and plot the de-pendence of lnW with respect to ln D . Using the same assumption used in de-

riving eq. (2.1) that expaW D∝ , determine the value of expa from the experiment.

[1.2 points]

The most widely accepted law that predicts the flow rate of grains through an orifice was proposed by Beverloo, which is modified from eq. (2.1) and has the form

( ) ,thraW C g D kdρ= − (2.2)

where d is the grain diameter and k is a dimensionless fitting parameter. According to the Beverloo law, the grains in fact do not flow through the whole orifice, but through an effective exit aperture whose diameter is given by ( ) D kd− .

Based on this law, determine the value of kd using linear regression from the da-ta you obtained. [0.8 points]


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Settling Rate of Sand

It is obvious that sand is a porous medium with different shapes and sizes of grains. In order to estimate the size of a sand grain, we can assume that the shape of the grain is spherical; however, the size of any individual grain might differ from that of the other grains. Despite this fact, we can still estimate the average size of sand par-ticles. In this experiment, you will observe the settling rate of sand particles in water and hence estimate the average radius of a sand grain.

Note: Use colored sand for this section. Plan your experiment very carefully before you begin your experiment, because you will not be able to recover the sand once it gets wet.

Determine the following quantities: Density of sand in air, mass of sand grains divided by the total volume (including

the volume of air inside the pores), Average density of sand grain, mass of sand grains divided by the total volume of sand grains. Density of sand in water, mass of sand grains divided by the total volume (includ-ing the volume of water inside the pores) [1.0 points]

Pour all the remaining colored sand into the graduated cylinder and fill it up with wa-ter. Close the cylinder using the plastic bag and the rubber band of the coloured sand, shake it well, and stand it in a vertical position. The sand will settle on the bot-tom with certain settling rate, which is the rate of the sediment height rises. Measure how fast the sand settles at various heights of the sediment and deter-

mine the settling rate of the sand within the region where the sand size is nearly homogenous. [1.0 points]

It is recommended to take measurements over the entire range of sediment volume in order to obtain the region where the sand size distribution is nearly uni-form/homogeneous. Derive an equation relating the sand settling rate and the density/ies of sand you

calculated before, taking note of the gravitational, Archimedean, and viscous forces. Give your estimation of the average size of the sand particles. [0.5 points]

The sand particle is assumed to be a spherical and any turbulent effect is neglected. You may also assume that a good shake causes the sand to be distributed uniformly. The formula for the viscous force is 6 ,viscous terminalF r vπ η= (2.3)

where r is the radius of the grain and η is the viscosity of water.


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Asteroid Impacts on Earths Surface When an asteroid impacts the surface of the Earth, it creates a large crater with circu-lar cross-section. The exact relation between the crater's dimensions, the physical properties of the impacting body, and the physical properties of the general area of impact is complicated and beyond the scope of this experiment. Nevertheless, there exists an empirical formula to estimate the diameter of an impact crater on the sur-face of the Earth based on experimental nuclear explosions at Yucca Flat, Nevada. From the experimental nuclear crater, the following empirical formula is obtained

1/

,af n

e

D c K Wσ

ρρ

=

(2.4)

where D is the diameter of the crater, fc is the so-called crater collapse factor

(whose value is 1 for craters m 3 k in diameter and 1.3 for craters m 4 k ), 1/0.074 km kiloto s nnK σ−= is an empirical constant, W is the kinetic energy of the as-

teroid just prior to impact, aρ is the mass density of the asteroid, eρ is the mean den-

sity of the target rocks at the point of impact and σ is a constant exponent. The di-ameter of the crater is defined by the peak of the "sand hill", not the outer most visi-ble boundary. The empirical formula above can be verified by modeling the collision using sand and glass balls. Note: Use the pile of sand in plastic basin Using the equipment provided, devise an experiment to determine the numerical

value of the constant exponent σ . After you make your measurement, you may need to loosen and flatten the surface of the sand again. If the sand is too com-pact it can affect your measurement result. [2.0 points]

Actually the σ of the typical asteroid is 25% bigger than the σ that we get from previous question. Estimate the diameter of a crater that would be caused by an iron meteorite of mass 710 kg drifting into Earth's path from very far away.

[0.5 points]

The students got detailed information about the use of the stopwatch in an appendix.


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Experimental Problem No. 2:

A Rotary Magnetic Drag System for Conductivity Measurement

Attention: Use the answer sheet provided to summarize your answer. Provide diagrams to help explaining your answer or your experiments, especially if your answer is not in English. Apparatus Rotary magnet assembly (x 1 pc) Copper plate (x 1 pc) "Black box" containing unknown metal X (x 1pc) Thin cardboard spacers (5 pcs) Thick cardboard spacers (2 pcs)

Power supply unit (x 1pc) Digital multimeter (DMM) – generic (x2 pcs). Digital multimeter (DMM) with frequency measurement (x1 pc). See Appendix A. Ruler (x 1 pc)

Figure 11: Experimental setup


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Important Experimental Data: Copper plate thickness: t = 0.6 mm Copper plate conductivity: σ = 6.0 x 107 (Ohm-m)-1 Metal X thickness: tX = 1.05 mm

— Introduction

Electrical conductivity measurement of a material is important for many applications such as metallurgy and semiconductor industries. Usually one measures conductivity (σ) (1/resistivity) by 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 meas-urement utilizing magnetic drag or braking effect that occurs between fast moving mag-nets and a metal sheet. If a magnet moves with velocity v parallel to the plane of a non-magnetic, conducting ma-terial with conductivity σ and thickness t, it will experience a magnetic braking effect also known as eddy current braking effect as shown below:

Figure 12: The magnetic braking effect of a moving magnet near a metal sheet


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The magnetic braking force (using a magnetic dipole model and a thin metal sheet ap-

proximation) is given as: ,MB m

vF td

ασ= − (2.1)

where α is the magnetic braking coefficient of the system which depends on magnetic moment of the magnet and magnetic permeability of the metal sheet, 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 thus it is called "magnetic braking". In this experiment we mount two strong magnetic pucks on a rotating disc driven by mo-tor as shown below. When the disc rotates and the metal plate is inserted underneath, the disc will slow down due to the magnetic braking effect. This effect can be exploited to measure the conductivity (or thickness) of a metal sheet.

Figure 13: Connection diagram of the rotary magnet setup

System Information: The motor is driven by a variable voltage power supply with coarse and fine control.

DMM (Digital Multimeter) #1 measures the motor voltage VM. DMM#2 measures the mo-tor current IM as a voltage across 1 Ω resistor. The Hall sensor serves as speed sensor. It provides a voltage pulse each time a magnet passes by. When the disc rotates, the Hall sensor frequency fH can be measured using DMM#3.


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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 13). Thick and thin card board spacers are provided to vary the distance d. 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 tX). See Section 1: “Important Experimental Data”

Attention: The disc can rotate up to fH = 200 Hz. In general, useful data can be ob-tained approximately < 120 Hz. Do not run the motor at maximum frequency (around 200 Hz) for too long. To save battery power, don't run the motor unnecessarily. Turn off the power if not in use.

— Experiment and Questions

This experiment is divided into four small sections: A. Introduction: Speed Sensor and Basic Operation [1.5 points] B. Basis System Characterization [2.0 points] C. Magnetic Braking Effect Characterization [4.5 points] D. Conductivity Measurement of an Unknown Metal X [2.0 points] --

A: Introduction: Speed Sensor and Basic Operation [1.5 points]

This section will help to familiarize yourself with the experimental setup. First we will try to understand the operation of the Hall effect sensor that measures the rotation frequen-cy. We will observe the voltage signal coming from the Hall sensor. Set the DMM #3 to DC Voltage mode. 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 near the sensor. Indicate the period of the waveform. [0.25 points]

Now switch DMM#3 to Frequency mode (Hz) to measure the Hall sensor frequency fH. 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 fre-quency reading fH increases with speed.


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Express the disc angular frequency ω (in rad/sec) in terms of Hall sensor frequency fH! [0.25 points]

Now insert the copper plate. Notice how the rotation slows down 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 below. Draw all the necessary electromagnet fields involved in this diagram in the answer sheet. [1.0 points]

B. Basic System Characterization [2 points] The motor has an internal series resistance RM which is not negligible in this experiment. RM is the sum of resistance of the rotor coil inside the motor. Therefore when the motor is driven by a voltage source, not all of the power is converted to kinetic or rotational ener-gy. A fraction of the energy is turned into heat due to RM. Thus the real motor can be modeled as an ideal motor (whose coil has no resistance) plus a series resistance RM as shown below.

Figure 14. The equivalent circuit of the motor


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Determine the internal series resistance of the motor RM to at least two significant figures! Note: The series resistance is very small. Direct resistance measurement using ohm meter mode of the DMMs (Digital Multimeters) does not have enough accuracy. Try other method. You don't have to change the experimental setup. [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 experi-ment. Under normal condition the motor should run smoothly, however at some frequen-cies the system vibrates strongly and become noisy and we want to avoid this. Do not insert the copper plate.

Increase the motor voltage gradually from 0 to near 200 Hz, record the Hall sensor frequency fH and the motor current IM (DMM#2). Plot both data with respect to voltage VM. Mark the range of frequencies where the system vibrates strongly and become noisy. Note that there is one dominant noisy region. [0.8 points]

What may cause this strong vibration/noise at certain frequencies? What anomaly do you observe in the motor current (IM) vs. voltage (VM) plot in this noisy region? [0.5 points]

C. Magnetic Braking Effect Characterization [4.5 points] Now we will study the characteristics of the magnetic braking effect with the copper plate provided. Here in section C and D, you can operate at low frequency (approximately fH < 120 Hz) and outside the dominant noisy region that you have identified in section B.

Derive expression for power dissipation PMB due to the magnetic braking force of the two magnets and the metal plate. Use (2.1) and express PMB in terms of the Hall sensor frequency fH. [0.5 points]

Perform experiments to verify the relationship of PMB and fH from your previous an-swer! [1.5 points]

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 Eq. (2.1) (due to magnetic dipole approximation) don't place the metal too close. Maintain sufficient distance, approximately d >8 mm.


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D. Conductivity Measurement of an Unknown Metal X [2.0 points] To obtain conductivity (assuming 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 large power factor m. In actual industrial ap-plication, it is difficult or even impossible to obtain an accurate distance d without making a contact to 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 un-known metal X inside a "black box" as shown below. The thickness is known: tX = 1.05 mm. However its exact location from the surface (∆d0) is not known. In fact, you do not need to know it to obtain the conductivity σ.

Figure 15: The cross section of the "black box" containing the unknown metal X.

Perform an experiment to determine the conductivity of the metal X! [2.0 points]

Note: Please don't open the "black box". It will invalidate your answer. The students got detailed information about the use of the multimeter in an appendix.

— 4. Result

WoPhO is an individual competition, that is why we will present only students’ results. There will be no compilation of results by countries. The best overall participant is Kexin Yi from China with a score of 35.1 out of a total maximum of 50.0. The best result for a theoretical examination is also achieved by Kexin Yi with a score of 22.75 out of a total maximum of 30.0. The best experimental result is achieved by Eugen Hruska from Ger-many with a score of 13.5 out of a total maximum of 20.0. The average score for each problem is very low due to many guest participants from the host country. We will present here only the average score of the medalists. The overall average score of medalists is 17.32, the theoretical average score is 9.59 and the experimental average score is 7.73. The lowest average for a theoretical problem is for problem no. 3, “Mirage”, with an aver-age of 1.5, followed by problem no. 1, “Motion of a Rolling Rod”, with an average of 2.44


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and the highest average is for problem no. 2, “Why Maglev Trains Levitate”, with an av-erage of 5.66. For the experimental problems, the average of problem no. 2 is 4.28, fol-lowed by problem no. 1 with an average of 3.46. In the following, we present the result of the 10 best students for their overall examina-tion, both the theoretical and the experimental examination.

Table 1: The best 10 participants overall

1 Yi, Kexin China 35.10 2 Ants Remm Estonia 27.63 3 Eugen Hruska Germany 27.48 4 Attila Szabo Hungary 25.35 5 Danila Parinov Russia 23.80 5 Christian George Emor Indonesia 23.78 7 Lin Sen Singapore 22.63 8 Jan Pulmann Slovakia 22.45 9 Li Kewei Singapore 20.75 10 Evan Laksono Indonesia 20.15

Table 2: The best 10 participants for the theoretical examination

1 Yi, Kexin China 22.75 2 Ants Remm Estonia 18.68 3 Attila Szabo Hungary 14.40 4 Eugen Hruska Germany 13.98 5 Danila Parinov Russia 13.90 5 Christian George Emor Indonesia 13.63 7 Evan Laksono Indonesia 12.90 8 Jan Pulmann Slovakia 12.50 9 Ivan Ivashkovskiy Russia 11.98 10 Samuel Wirajaya Indonesia 11.83


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Table 3: The best 10 participants for the experimental examination

1 Eugen Hruska Germany 13.50 2 Yi, Kexin China 12.35 3 Lin Sen Singapore 11.40 4 Li Kewei Singapore 11.15 5 Attila Szabo Hungary 10.95 5 Christian George Emor Indonesia 10.15 7 Jan Pulmann Slovakia 9.95 8 Danila Parinov Russia 9.90 9 Wojciech Tarnowski Poland 9.60 10 Momchil Molnar Bulgaria 9.55

The authors would like to make two observations. First, WoPhO has been promoted as an opportunity for any student to challenge the win-ners of APhO and IPhO. However, all the gold medalists at the first WoPhO were still gold medalists of either the competition year APhO or IPhO, with only 2 exceptions. One of the two students, however, was a former gold medalist from the previous year IPhO and another one will be the absolute winner of the next IPhO. Second, the average score even for the medalists of the first final round of WoPhO was quite low. The problems were difficult and both the academic committee and the stu-dents acknowledged this fact. However, WoPhO is also a competition for the winners of the previous Olympiads and the level of difficulty of the problems in that regard can be justified.

physics competitions vol. 14 no 2 2012 the final...

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