[User]
USAPhO 2020, problem A3. An unusual problem that tests your understanding
    of momentarily comoving frames, and higher-dimensional Lorentz transformations. As a warning,
    this question requires you to make an unstated assumption. The fact that uniformly moving clocks
    have their time dilated by a factor of γ follows directly from the postulates of special relativity.
    But here you’ll have to assume this also holds for accelerating clocks, even though clocks can tell
    if they’re accelerating, and may tick differently. This is called the clock hypothesis. For example,
    on a roller coaster, a pendulum clock doesn’t obey the clock hypothesis, but a quartz watch does.
    Also, the solution is a bit misleading, so don’t worry if you thought about the problem differently
    as long as you got the same final answers.
[3] Problem 27 (Morin 11.26). The following problem is called Bell’s spaceship paradox. It caused a
    stir at CERN when many particle physicists could not agree on the answer.
        Two spaceships float in space and are at rest relative to each other. They are connected by a
    string. The string is strong, but it cannot withstand an arbitrary amount of stretching.

    At a given instant, the spaceships simultaneously (with respect to their initial inertial frame) start
    accelerating in the same direction along the line between them, with the same constant proper
    acceleration. In other words, assume they bought identical engines from the same store, and they

                    01h
    put them on the same setting. Will the string eventually break?

Solution:
The two conflicting ideas is that in the lab frame, length contraction would indicate
    that the distance between the two rockets is γL in the co-moving frame, indicating that the string
    stretches and breaks. The other idea is that in the rocket frame, they both accelerate with the same
    proper acceleration, and the relative velocity stays as 0.
       The correct answer is that the string will eventually break. The second line of reasoning fails
    because of the relativity of simultaneity. We can consider, at each moment, the inertial frame
    comoving with the rear rocket, whose axes are as shown.

       Over time, the x′ axis of this set of inertial frames tilts upward, so the rear rocket sees the front
   rocket accelerating faster, and hence moving away.
       If this isn’t clear from the diagram, one can consider discretizing the acceleration, i.e. converting
   it into a series of rocket pulses. Since the clocks of the rockets are synchronized, the first pulses
   are simultaneous according to both rockets. But because of the relativity of simultaneity, within
   the frame moving with the rear rocket after the first pulse, the front rocket does the second pulse
   earlier, and hence starts to move away. (For a neat visual explanation, see this video.)


[User]
01T IPhO 1995, problem 2. Refraction in the presence of a linearly varying wave
    speed. (This is a classic setup with a neat solution, also featured in IPhO 1974, problem 2.)

                    01h

Solution:
01TIPhO 1995, problem 2. Refraction in the presence of a linearly varying wave
    speed. (This is a classic setup with a neat solution, also featured in IPhO 1974, problem 2.)

Kevin Zhou Physics Olympiad Handouts — educational use, attribute to Kevin Zhou (knzhou.github.io). LICENSE FLAG: must reconfirm permissions before public release.


[User]
An electron beam is accelerated from rest by applying an electric field E for a time
    t, and subsequently guided by magnetic fields. These magnetic fields are produced with a series of
    coils, which carry currents Ii .
        Now suppose the apparatus is repurposed to shoot proton beams. Suppose a proton beam is
    accelerated from rest by applying an electric field E for a time t (in the opposite direction). Let an
    electron have mass m and a proton have mass M .

      (a) Find the currents Ii needed so that the proton follows the same trajectory the electron did,
          assuming V is small enough that both the electron and proton are nonrelativistic.

      (b) How does the answer change if relativistic corrections are accounted for?

Solution:
(a) The electron and proton have the same momentum p, and we have
                                                   dp
                                                      = qvB ∼ qvI
                                                   dt
          since B ∝ I. Now, the magnetic field can only rotate the particle’s momentum. Suppose at
          some moment it is curving in a trajectory with radius of curvature r, and speed v. Then it
          has instantaneous angular velocity ω = v/r along the circle tangent to its trajectory, so
                                                       dp
                                                          = ωp.
                                                       dt
          Hence we have
                                                           v
                                                       qvI ∼ p
                                                           r
          and since r is the same for both the electron and proton, we have the scaling
                                                          p
                                                      I∝ .
                                                          q
          By construction, the electron and the proton have the same momentum p, so we have I ∝ 1/q.
          That is, to accommodate the proton, we just have the flip the sign of the currents.

     (b) Every step in the solution to part (a) still works with relativity accounted for (the change of
         p = mv to p = γmv doesn’t matter, because we never used p = mv), so the answer is the

                   01h
         same: we just flip the currents.


[User]
Another type of boundary condition is the “soft” boundary condition, which requires
    dy/dx = 0 at x = 0. Show that waves are reflected from this boundary but not flipped.

Solution:
Let the reflected wave again be a general wave traveling backward,

                                           yr (x, t) = g(−kx − ωt).

    The boundary condition is
                      ∂
                         (f (kx − ωt) + g(−kx − ωt)) |x=0 = kf ′ (−ωt) − kg ′ (−ωt) = 0
                      ∂x
    which tells us that f = g up to a constant. Of course, that constant is just the initial height of the
    string at x = 0, which we set to zero. Thus, f = g, so the wave is reflected without a sign flip.


[User]
Question A4
A heat engine consists of a moveable piston in a vertical cylinder. The piston is held in place by
a removable weight placed on top of the piston, but piston stops prevent the piston from sinking
below a certain point. The mass of the piston is m = 40.0 kg, the cross sectional area of the piston
is A = 100 cm2 , and the weight placed on the piston has a mass of m = 120.0 kg.
    Assume that the region around the cylinder and piston is a vacuum, so you don’t need to worry
about external atmospheric pressure.

   • At point A the cylinder volume V0 is completely filled with liquid water at a temperature
     T0 = 320 K and a pressure Pmin that would be just sufficient to lift the piston alone, except
     the piston has the additional weight placed on top.

   • Heat energy is added to the water by placing the entire cylinder in a hot bath.

   • At point B the piston and weight begins to rise.

   • At point C the volume of the cylinder reaches Vmax and the temperature reaches Tmax . The
     heat source is removed; the piston stops rising and is locked in place.

   • Heat energy is now removed from the water by placing the entire cylinder in a cold bath.

   • At point D the pressure in the cylinder returns to Pmin . The added weight is removed; the
     piston is unlocked and begins to move down.

   • The cylinder volume returns to V0 . The cylinder is removed from the cold bath, the weight
     is placed back on top of the piston, and the cycle repeats.

   Because the liquid water can change to gas, there are several important events that take place

   • At point W the liquid begins changing to gas.

   • At point X all of the liquid has changed to gas. This occurs at the same point as point C
     described above.

   • At point Y the gas begins to change back into liquid.

   • At point Z all of the gas has changed back into liquid.

   When in the liquid state you need to know that for water kept at constant volume, a change in
temperature ∆T is related to a change in pressure ∆P according to

                                       ∆P ≈ (106 Pa/K)∆T

   When in the gas state you should assume that water behaves like an ideal gas.

                   Copyright c 2015 American Association of Physics Teachers

2015 USA Physics Olympiad Exam                              Part A                               7

   Of relevance to this question is the pressure/temperature phase plot for water, showing the re-
gions where water exists in liquid form or gaseous form. The curve shows the coexistence condition,
where water can exist simultaneously as gas or liquid.

                                      240

                                      200

                                      160

                     Pressure (kPa)
                                      120
                                                        Liquid Region
                                       80

                                       40
                                                                        Gas Region
                                       0
                                            280   300    320   340     360   380     400
                                                          Temperature (K)

   The following graphs should be drawn on the answer sheet provided.

  a. Sketch a PT diagram for this cycle on the answer sheet. The coexistence curve for the
     liquid/gas state is shown. Clearly and accurately label the locations of points B through D
     and W through Z on this cycle.

  b. Sketch a PV diagram for this cycle on the answer sheet. You should estimate a reasonable
     value for Vmax , note the scale is logarithmic. Clearly and accurately label the locations of
     points B through D on this cycle. Provide reasonable approximate locations for points W
     through Z on this cycle.

                   Copyright c 2015 American Association of Physics Teachers

2015 USA Physics Olympiad Exam          Part A                             8

   STOP: Do Not Continue to Part B

If there is still time remaining for Part A, you should review your work for
    Part A, but do not continue to Part B until instructed by your exam
                                  supervisor.

               Copyright c 2015 American Association of Physics Teachers

2015 USA Physics Olympiad Exam                Part B                                               9

Part B

Solution:
Question A4
A heat engine consists of a moveable piston in a vertical cylinder. The piston is held in place by
a removable weight placed on top of the piston, but piston stops prevent the piston from sinking
below a certain point. The mass of the piston is m = 40.0 kg, the cross sectional area of the piston
is A = 100 cm2 , and the weight placed on the piston has a mass of m = 120.0 kg.
    Assume that the region around the cylinder and piston is a vacuum, so you don’t need to worry
about external atmospheric pressure.

   • At point A the cylinder volume V0 is completely filled with liquid water at a temperature
     T0 = 320 K and a pressure Pmin that would be just sufficient to lift the piston alone, except
     the piston has the additional weight placed on top.

   • Heat energy is added to the water by placing the entire cylinder in a hot bath.

   • At point B the piston and weight begins to rise.

   • At point C the volume of the cylinder reaches Vmax and the temperature reaches 400 K. The
     heat source is removed; the piston stops rising and is locked in place.

   • Heat energy is now removed from the water by placing the entire cylinder in a cold bath.

   • At point D the pressure in the cylinder returns to Pmin . The added weight is removed; the
     piston is unlocked and begins to move down.

   • The cylinder volume returns to V0 . The cylinder is removed from the cold bath, the weight
     is placed back on top of the piston, and the cycle repeats.

   Because the liquid water can change to gas, there are several important events that take place

   • At point W the liquid begins changing to gas.

   • At point X all of the liquid has changed to gas. This may not occur the same as point C
     described above.

   • At point Y the gas begins to change back into liquid.

   • At point Z all of the gas has changed back into liquid.

   When in the liquid state you need to know that for water kept at constant volume, a change in
temperature ∆T is related to a change in pressure ∆P according to

                                       ∆P ≈ (106 Pa/K)∆T

   When in the gas state you should assume that water behaves like an ideal gas.

                   Copyright c 2015 American Association of Physics Teachers

2015 USA Physics Olympiad Exam                                  Part A                                       10

   Of relevance to this question is the pressure/temperature phase plot for water, showing the re-
gions where water exists in liquid form or gaseous form. The curve shows the coexistence condition,
where water can exist simultaneously as gas or liquid.
                                      240

                                      200

                                      160

                     Pressure (kPa)
                                      120
                                                           Liquid Region
                                       80

                                       40
                                                                                Gas Region
                                       0
                                            280      300     320   340     360                 380     400
                                                              Temperature (K)

   The following graphs should be drawn on the answer sheet provided.

                                                             Solution
   Before we get started solving the problem, let’s say a bit about where the data in the problem
comes from. We are using the Magnus form to approximate the coexistence curve,
                                                  P = (610.94 Pa)e17.625/(1+243.04/T )
where T is measured in centigrade. This is closely related to the result that can be derived from
the Clausius-Clapeyron equation for ideal gases,

                                                                      L       T −T0
                                                           P = P0 e   R        T T0

where we assume the temperature is low compared to the critical temperature and the latent heat
L is a constant.
    To get ∆P/∆T , we used the cyclic chain rule

                                 ∂P       ∂V      ∂T
                                                         = −1
                                 ∂T V ∂P T ∂V P
where subscripts indicate what is being held constant. Dropping those for convenience,
                                                     (6 × 10−4 K−1 )

              ∂P           ∂P      1 ∂V      βV T
                   = −V                   =       ≈                   ≈ 106 Pa/K.
              ∂T           ∂V     V ∂T       βP V   (5 × 10−10 Pa−1 )
The specific value is not important; the point is that a very small change in the temperature of the
liquid in a fixed volume will result in a very large change in the pressure.

                   Copyright c 2015 American Association of Physics Teachers

2015 USA Physics Olympiad Exam                               Part A                            11

  a. Sketch a PT diagram for this cycle on the answer sheet. The coexistence curve for the
     liquid/gas state is shown. Clearly and accurately label the locations of points B through D
     and W through Z on this cycle.

  b. Sketch a PV diagram for this cycle on the answer sheet. You should estimate a reasonable
     value for Vmax , note the scale is logarithmic. Clearly and accurately label the locations of
     points B through D on this cycle. Provide reasonable approximate locations for points W
     through Z on this cycle.

                                                           Solution
The correct graphs are shown below.

                                     240

                                     200

                                                       B                C, W, X, Y
                                     160

                    Pressure (kPa)
                                     120

                                      80

                                                       A
                                      40
                                                                      D, Z
                                      0
                                           280   300       320   340     360   380   400
                                                            Temperature (K)

                  Copyright c 2015 American Association of Physics Teachers

2015 USA Physics Olympiad Exam                                Part A                               12

                                      240

                                      200

                                                B       W                     C, X, Y
                                      160

                     Pressure (kPa)
                                      120

                                      80

                                                A       Z                     D
                                      40

                                       0
                                            0       1      10    102     103     104    105
                                                        Volume (in terms of V0 )

    We start by computing pressures. The minimum pressure is attained when only the piston is
to be lifted, so
                               F     mg   (40 kg)(10 m/s2 )
                        Pmin =    =     =                   = 40 kPa.
                               A     A        (0.01 m2 )
The maximum pressure is attained when lifting the piston with extra weight,

                                                F   mg   (160 kg)(10 m/s2 )
                                      Pmax =      =    =                    = 160 kPa.
                                                A   A        (0.01 m2 )

Point A is clearly at (Pmin , T0 ) on the PT graph.
    For liquid water a small temperature increase results in a large pressure increase, so point B is
effectively at the same temperature as point A. Process A → B is therefore essentially isothermal,
and it is also a constant volume process.
    Afterwards, the pressure is sufficient to lift the piston and weight, so the volume expands at
constant pressure for the process B → C. However, liquid cannot change to gas until we reach the
coexistence curve. This defines the location of point W. On the PT graph we are “stuck” on the
coexistence curve until all the liquid has changed into gas, so X and C are also at the same point.
    Upon reaching C the piston is locked in place, fixing the volume, and the cylinder is allowed
to cool. To figure out what happens, suppose water vapor obeyed the ideal gas law P V = nRT .
For a constant volume process, T ∝ P , so the path would be a straight line towards the origin of
the P T diagram. This isn’t what happens here, because we run into the coexistence curve, and
the pressure is decreased by some of the gas condensing to liquid. Thus during the entire C → D
process, we follow the coexistence curve downward; it is impossible to cross it until all of the vapor
condenses.
    The next process C → D is constant volume, but not isothermal. On the PT graph we follow
the coexistence curve to the minimum pressure, at which time the piston is freed and allowed to

                   Copyright c 2015 American Association of Physics Teachers

2015 USA Physics Olympiad Exam                Part A                                             13

lower at constant pressure. Since Z is the point where all of the gas has changed to liquid, it must
be on the coexistence curve. On the PV diagram it is just to the right of point A.
    There are a few numbers needed on the PV diagram that are not given in the problem. You
should know from everyday experience that liquid water will only expand slightly when heated
over this temperature range, so point W must be very close to B on the PV diagram. You might
also know that the density of liquid water is about 2, 000 times the density of water vapor in these
conditions. In terms of grading policy, the points C, X, Y, and D may have volumes in the range
[1000V0 , 5000V0 ] on the PV diagram for full credit, while points W and Z may have volumes in the
range [V0 , 2V0 ] for full credit and [2V0 , 5V0 ] for partial credit.

                   Copyright c 2015 American Association of Physics Teachers

2015 USA Physics Olympiad Exam          Part A                             14

   STOP: Do Not Continue to Part B

If there is still time remaining for Part A, you should review your work for
    Part A, but do not continue to Part B until instructed by your exam
                                  supervisor.

               Copyright c 2015 American Association of Physics Teachers

2015 USA Physics Olympiad Exam                Part B                                              15

Part B

USAPhO 2015 Question A4


[User]
A point source of light L emitting a single wavelength λ is situated a distance d
    from an ideal mirror, at z = 0. A screen stands at the end of the mirror at distance D ≫ d from L.

    Find the relative intensity of light on the screen as a function of z. (This setup is known as Lloyd’s
    mirror.)

Solution:
A point source of light L emitting a single wavelength λ is situated a distance d
    from an ideal mirror, at z = 0. A screen stands at the end of the mirror at distance D ≫ d from L.

    Find the relative intensity of light on the screen as a function of z. (This setup is known as Lloyd’s
    mirror.)

    Answer. I(z) ∝ sin2 (2πdz/Dλ)


[User]
Consider a long cylindrical capacitor whose surfaces are concentric cylinders of radius rin and rout .
The capacitor is filled with a gas which will break down if it experiences an electric field greater
than E0 . If the outer radius is fixed, which should the inner radius be in order to maximize the
voltage V to which the capacitor can be charged, without breakdown?

Solution:
Consider a long cylindrical capacitor whose surfaces are concentric cylinders of radius rin and rout .
The capacitor is filled with a gas which will break down if it experiences an electric field greater
than E0 . If the outer radius is fixed, which should the inner radius be in order to maximize the
voltage V to which the capacitor can be charged, without breakdown?
Solution. Let the charge per length on the cylinders be ±λ. By Gauss’s law, the electric field is
                                                     λ
                                             E=          .
                                                   2πrϵ0
Thus, the voltage drop is
                                             λ
                                     V =          log(rout /rin ).
                                            2πϵ0
On the other hand, the highest electric field is right at the inner cylinder, so
                                                     λ
                                           E0 =            .
                                                  2πrin ϵ0
Thus, we fix λ ∝ rin , which means
                         V ∝ rin log(rout /rin ) ∝ x log(1/x),   x = rin /rout .
Setting the derivative to zero,
                                           log(1/x) − 1 = 0
which implies x = 1/e, or rin = rout /e.


[User]
3. Chandrayaan-3
                                        CS
  On July 14, 2023, India’s lunar mission satellite, Chandrayaan-3, was successfully launched by
  the Indian Space Research Organization (ISRO). Chandrayaan-3 (mass m =3900 kg) was taken
  to the Moon through a series of Earth Bound Manoeuvres (elliptical) orbits (EBNs) as depicted
  in the ﬁgure below. In this problem, we will explore the physics governing some part of its

                  HB
  journey, employing a simpliﬁed model. For all parts of this problem except part (f), we consider
  Chandrayaan-3 to be moving only under the inﬂuence of Earth’s gravity (a central force).

   (a) [6 marks] Upon launch, Chandrayaan-3 entered an elliptical orbit around Earth, with Earth
       at one of the foci (E) as shown below. The points P and A are the perigee (nearest point
       from the Earth) and apogee (farthest point from the Earth), respectively. We introduce the
       polar coordinate system (r,θ), where ⃗r is the vector from the centre of the Earth (origin) to
       the satellite, and θ is the angle that ⃗r makes with the major axis (PA= 2a). The directions
       of unit vectors r̂ and θ̂ are shown in the ﬁgure.

Page 5       INPhO 2024                     Questions

                                                                 θ̂
                                                                            r̂

                                                     r
                                                            ⃗r
                                                 θ
                                  P                                              A
                                        E

         The equation of the ellipse can be written in polar coordinates as
                                                              r0
                                                 r=
                                                         (1 − e cos θ)
         where e is eccentricity of the orbit (0 < e < 1) and r0 is called the latus rectum.
         The velocity ⃗v of the satellite in polar coordinates can be written as

                                            ⃗v = vr r̂ + vt θ̂ = ṙr̂ + rθ̇θ̂

         where vr = ṙ is the “radial” speed and vt = rθ̇ is the “tangential” speed.
         Make schematic plots of the speeds vr and vt as functions of θ over one full orbit. Mark any

                                                          E
         signiﬁcant points in the plots in terms of a, e, and other variables.

                                  CS
            HB

Page 6      INPhO 2024                     Questions          Last four digits of Roll No.:

   (b) [1.5 marks] Obtain an expression for the total energy (E) of the orbiting satellite in terms
       of a and other constants.
   (c) [1 marks] Plot the kinetic energy (KE) of the satellite as a function of θ over one full orbit.
       Mark any signiﬁcant points in terms of a, e, and other variables.
   (d) [1.5 marks] The perigee and apogee of the elliptical orbit in part (a) are 200 km and 36500
       km, respectively. It is generally described as a (200×36500) km orbit. Here the distances
       are deﬁned from the surface of the Earth. Calculate the period of rotation T (in hr) of
       Chandrayaan-3 in this orbit.
   (e) [2.5 marks] To move Chandrayaan-3 from the ﬁrst orbit (in part (d)) to another elliptical
       orbit EBN-1, an instantaneous boost was applied at perigee by changing the velocity by ∆v,
       without altering the direction. This changed the apogee to 41800 km above Earth’s surface
       while keeping the perigee unchanged. Calculate ∆v.
    (f) [1.5 marks] After a series of manoeuvres, Chandrayaan-3 was placed in an elliptical orbit
        of (100 × 1437) km around the Moon. Here, the distances are calculated from the surface
        of the Moon. Calculate the change in velocity ∆v ′ , applied at the perigee, that is required
        to bring Chandrayaan-3 from this elliptical orbit to a circular orbit at a distance of 100 km
        from the surface of the Moon. For this part, assume that Chandrayaan-3 is only under the
        inﬂuence of the Moon’s gravitational ﬁeld.

Solution:
3. Chandrayaan-3
  On July 14, 2023, India’s lunar mission satellite, Chandrayaan-3, was successfully launched by
  the Indian Space Research Organization (ISRO). Chandrayaan-3 (mass m =3900 kg) was taken
  to the Moon through a series of Earth Bound Manoeuvres (elliptical) orbits (EBNs) as depicted
  in the ﬁgure below. In this problem, we will explore the physics governing some part of its
  journey, employing a simpliﬁed model. For all parts of this problem except part (f), we consider
  Chandrayaan-3 to be moving only under the inﬂuence of Earth’s gravity (a central force).

                                                                        E
                                        CS
                  HB
   (a) [6 marks] Upon launch, Chandrayaan-3 entered an elliptical orbit around Earth, with Earth
       at one of the foci (E) as shown below. The points P and A are the perigee (nearest point
       from the Earth) and apogee (farthest point from the Earth), respectively. We introduce the
       polar coordinate system (r,θ), where ⃗r is the vector from the centre of the Earth (origin) to
       the satellite, and θ is the angle that ⃗r makes with the major axis (PA= 2a). The directions
       of unit vectors r̂ and θ̂ are shown in the ﬁgure.
                                                                   θ̂
                                                                             r̂

                                                     r
                                                             ⃗r
                                                 θ
                                 P                                                        A
                                       E

         The equation of the ellipse can be written in polar coordinates as
                                                              r0
                                                 r=
                                                         (1 − e cos θ)

Page 9       INPhO 2024                    Questions

         where e is eccentricity of the orbit (0 < e < 1) and r0 is called the latus rectum.
         The velocity ⃗v of the satellite in polar coordinates can be written as

                                           ⃗v = vr r̂ + vt θ̂ = ṙr̂ + rθ̇θ̂

         where vr = ṙ is the “radial” speed and vt = rθ̇ is the “tangential” speed.
         Make schematic plots of the speeds vr and vt as functions of θ over one full orbit. Mark any
         signiﬁcant points in the plots in terms of a, e, and other variables.

                                                        E
                                  CS
            HB

Page 10      INPhO 2024                     Questions               Last four digits of Roll No.:

 Solution: It is given that
                                                               r0
                                                   r=                                                (3.1)
                                                          (1 − e cos θ)
                                                           r0 e sin θθ̇
                                       ∴ vr = ṙ = −                                                 (3.2)
                                                         (1 − e cos θ)2
 Since the force is central, angular momentum l is conserved. The conserved angular momentum
 is given by

                                                l = mr2 θ̇                                           (3.3)
                                                      l
                                             =⇒ θ̇ =                                                 (3.4)
                                                     mr2
 Substituting above equation and expression of r in vr , after simplifying, we get
                                                         el sin θ
                                             vr = −                                                  (3.5)
                                                          mr0
We know that
                                                          l
                                                  vt =                                               (3.6)
                                                         mr
 Substituting the value of r, we get

                                           vt =
                                                  l(1 − e cos θ)
                                                       mr0
                                                                    E                                (3.7)

                                         CS
 From above equation, the value of vt is maximum, when cos θ is minimum. i.e. θ = −π. The
 maximum value of vt

                                            vtmax =
                                                         l(1 + e)
                                                           mr0
 From Eq. (3.7), vt is minimum, when cos θ is maximum. i.e θ = 0. The minimum value of vt
                                                                                (3.8)

                 HB                         vtmin =
                                                       l(1 − e)
                                                         mr0
 Since the velocities at P and A are purely tangential, and P is closer to earth as compared to A,
 hence the tangential speed is maximum at P. Let the velocity of the satellite in the given orbit at
 P be vp and velocity at A be va . Conservation of energy gives:
                                                                                (3.9)

                                 1        GM m  1       GM m
                                   mv 2 −      = mvp2 −                                             (3.10)
                                 2 a       ra   2        rp

 Conservation of angular momentum gives

                                            mva ra = mvp rp                                         (3.11)
                                                        ra
                                                vp = va                                             (3.12)
                                                        rp

 Substituting the above equation into the energy conservation equation, we get
                                            √
                                               GM (1 + e)
                                       vp =                                                         (3.13)
                                                a (1 − e)

 Similarly
                                                  √
                                                      GM (1 − e)
                                          va =                                                      (3.14)
                                                       a (1 + e)

Page 11       INPhO 2024                      Questions

 Hence the conserved angular momentum can be written as

                                             l = mva ra = mva ra                               (3.15)

 On simpliﬁcation, we get

                                                       √
                                                             GM
                                             l = mro                                           (3.16)
                                                           a(1 − e2 )

 vr and vt can also be written as

                                                      √
                                                            GM
                                          vr = −e                    sin θ                     (3.17)
                                                          a(1 − e2 )
                                             √
                                      vt =
                                                   GM
                                                 a(1 − e2 )
                                                            E
                                                            (1 − e cos θ)                      (3.18)

                                    CS
 Similarly vrmax , vrmin , vtmax and vtmin can also be written as

                    vrmax = e
                                √
                                 GM
                               a(1 − e2 )
                               √
                                            at θ = 3π/2                                        (3.19)

            HB      vrmin = −e

                     max
                    vt    =

                    vtmin =
                            √

                            √
                                   GM
                                 a(1 − e2 )

                                GM
                              a(1 − e2 )

                                GM
                                              at θ = π/2

                                         (1 + e) =

                                         (1 − e) =
                                                   √

                                                   √
                                                     GM (1 + e)
                                                      a(1 − e)

                                                     GM (1 − e)
                                                                             at θ = π

                                                                             at θ = 0 and 2π
                                                                                (3.20)

                                                                                (3.21)

                                                                                (3.22)
                              a(1 − e )
                                     2                a(1 + e)

Page 12      INPhO 2024                     Questions            Last four digits of Roll No.:

   (b) [1.5 marks] Obtain an expression for the total energy (E) of the orbiting satellite in terms
       of a and other constants.

 Solution: Let
                                             va
                                             rp
                                                =
                                                  vp
                                                  ra
                                                     =C     E                                    (3.23)

                                       CS
 Putting Eq. (3.23) in Eq. (3.10), we get

                              1
                              2

 solving above equation, we get
                                mC 2 rp2 −
                                           GM m
                                            ra
                                                 1
                                                = mC 2 ra2 −
                                                 2
                                                             GM m
                                                              rp
                                                                                (3.24)

                 HB                     C2 =
                                                   2GM
                                               (rp + ra )ra rp

Again using Eq. (3.23) and value of C 2 from above equation in LHS of the Eq. (3.24), we get

                                    1
                                  E= m
                                          2GM rp2
                                    2 ra rp (ra + ra )
                                                       −
                                                         GM m
                                                          ra
                                                                                (3.25)

                                                                                (3.26)

 Solving above equation, and substituting ra + rp = 2a, we get

                                                   GM m
                                             E=−                                                 (3.27)
                                                    2a

   (c) [1 marks] Plot the kinetic energy (KE) of the satellite as a function of θ over one full orbit.
       Mark any signiﬁcant points in terms of a, e, and other variables.

 Solution:
                                                 1
                                             KE = mv 2                                           (3.28)
                                                 2
where v 2 = vr2 + vt2 . Substituting the value of vr and vt in the above equation, we get

                                         GM m(1 + e2 − 2e cos θ)
                                  KE =                                                           (3.29)
                                              2a(1 − e2 )

Page 13      INPhO 2024                  Questions

          For θ = 0, KE = KEmin = GM m(1−e)                            GM m(1+e)
                                   2a(1+e) , and for θ = π KE = KEmax = 2a(1−e)

   (d) [1.5 marks] The perigee and apogee of the elliptical orbit in part (a) are 200 km and 36500
       km, respectively. It is generally described as a (200×36500) km orbit. Here the distances
       are deﬁned from the surface of the Earth. Calculate the period of rotation T (in hr) of
       Chandrayaan-3 in this orbit.

 Solution: The period of the orbit is given by
                                                    E
            r +r
                                   CS      T2 =
                                                  4π 2 3
                                                  GM
                                                      a                                    (3.30)

where a = p 2 a , where rp is perigee distance given by rp = 200km + RE and ra is apogee distance
given by ra = 36500km + RE and i.e. a = 6371 + (200+36500)km
                               √
                                                          2     = 24721km

            HB           T =
                                         4 × π2
                               6.67 × 10−11 × 5.972 × 1024
                           = 10.75 hr
                                                           247213 × 109                    (3.31)

                                                                                (3.32)

   (e) [2.5 marks] To move Chandrayaan-3 from the ﬁrst orbit (in part (d)) to another elliptical
       orbit EBN-1, an instantaneous boost was applied at perigee by changing the velocity by ∆v,
       without altering the direction. This changed the apogee to 41800 km above Earth’s surface
       while keeping the perigee unchanged. Calculate ∆v.

 Solution: since the velocities at P and A are purely tangential, and P is closer to earth as
 compared to A, hence the tangential speed is maximum at P. Let the velocity of the satellite in
 the given orbit at P be vp and velocity at A be va . Conservation of energy gives:

                                   1        GM m  1       GM m
                                     mva2 −      = mvp2 −                                  (3.33)
                                   2         ra   2        rp

 Conservation of angular momentum gives

                                          mva ra = mvp rp                                  (3.34)
                                                      ra
                                              vp = va                                      (3.35)
                                                      rp

Page 14       INPhO 2024                   Questions                   Last four digits of Roll No.:

 Substituting the above equation into the energy conservation equation, we get
                                            √
                                                 2GM ra
                                       vp =                                                            (3.36)
                                               rp (rp + ra )
                                            √
                                               GM ra
                                       vp =                                                            (3.37)
                                                 arp
                                            √
                                               GM (1 + e)
                                       vp =                                                            (3.38)
                                                 a (1 − e)

 Similarly
                                                  √
                                                        2GM rp
                                         va =                                                          (3.39)
                                                      ra (rp + ra )
                                                  √
                                                      GM rp
                                         va =                                                          (3.40)
                                                       ara
                                                  √
                                                      GM (1 − e)
                                         va =                                                          (3.41)
                                                       a (1 + e)

 From Eq. (3.36)

                                          vp =
                                                  √
                                                        2GM ra    E
                                                      rp (rp + ra )
                                                                                (3.42)

                                        CSvp = 10.251 km/s

 In the new orbit, the perigee distance is kept the same, and the apogee distance changed to 41800
 km, hence ra = 41800 + RE = 48171km and rp remains as it is. Hence, the new velocity at the
 perigee distance is vp′ Using the expression Eq. (3.36), the vp′ can be written as
                                                  √
                                                                                (3.43)

                   HB                     vp′ =
                                                        2GM ra
                                                      rp (ra′ + rp )
                                          vp′ = 10.327 km/s

 Now the boost required ∆v at this position would be

                                          ∆v = vp′ − vp
                                                                                (3.44)

                                                                                (3.45)

                                                                                (3.46)
                                          ∆v = 0.076 km/s                                              (3.47)

    (f) [1.5 marks] After a series of manoeuvres, Chandrayaan-3 was placed in an elliptical orbit
        of (100 × 1437) km around the Moon. Here, the distances are calculated from the surface
        of the Moon. Calculate the change in velocity ∆v ′ , applied at the perigee, that is required
        to bring Chandrayaan-3 from this elliptical orbit to a circular orbit at a distance of 100 km
        from the surface of the Moon. For this part, assume that Chandrayaan-3 is only under the
        inﬂuence of the Moon’s gravitational ﬁeld.

 Solution: Let the apogee distance be rma = Rm +1437 km and perigee distance be rmp = Rm +100
 km The velocity of the satellite at perigee, when it is in the elliptic orbit, is vmp and the velocity

Page 15     INPhO 2024                     Questions

when it is in the circular orbit is vm we know that
                                            √
                                                2GMM rma
                                      vmp =                                                  (3.48)
                                              rmp (rma + rmp )
                                    vmp = 1.838 km/s                                         (3.49)

 Similarly here apogee distance be rma = Rm + 100 km and perigee distance be rmp = Rm + 100
 km. Let rm = rma = rmp
                                           √
                                              2GMm rm
                                      vm =                                            (3.50)
                                             rm (rm + rm )
                                      vm = 1.633 km/s                                        (3.51)
                                                                                (3.52)

The boost required ∆v ′

                                          ∆v ′ = vm − vmp                                    (3.53)
                                            ′
                                          ∆v = −0.205 km/s                                   (3.54)

                                                     E

INPhO 2024 Problem 3


[User]
39th International Physics Olympiad - Hanoi - Vietnam - 2008
                                     Theoretical Problem No. 1

                 WATER-POWERED RICE-POUNDING MORTAR

A. Introduction
    Rice is the main staple food of most people in Vietnam. To make white rice from
paddy rice, one needs separate of the husk (a process called "hulling") and separate the
bran layer ("milling"). The hilly parts of northern Vietnam are abundant with water
streams, and people living there use water-powered rice-pounding mortar for bran layer
separation. Figure 1 shows one of such mortars., Figure 2 shows how it works.

B. Design and operation
    1. Design.
    The rice-pounding mortar shown in Figure 1 has the following parts:
   The mortar, basically a wooden container for rice.
    The lever, which is a tree trunk with one larger end and one smaller end. It 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.
    2. Modes of operation
    The mortar has two modes.
    Working mode. In this mode, the mortar goes through an operation cycle illustrated in
Figure 2.
    The rice-pounding function comes from the work that is transferred from the pestle to
the rice during stage f) of Figure 2. If, for some reason, the pestle never touches the rice,
we say that the mortar is not working.
    Rest mode with the lever lifted up. During stage c) of the operation cycle (Figure 2),
as the tilt angle α increases, the amount of water in the bucket decreases. At one
particular moment in time, the amount of water is just enough to counterbalance the
weight of the lever. Denote the tilting angle at this instant by β . If the lever is kept at
angle β and the initial angular velocity is zero, then the lever will remain at this
position forever. This is the rest mode with the lever lifted up. The stability of this
position depends on the flow rate of water into the bucket, Φ . If Φ exceeds some

value Φ 2 , then this rest mode is stable, and the mortar cannot be in the working mode.

In other words, Φ 2 is the minimal flow rate for the mortar not to work.

                                                                                1

39th International Physics Olympiad - Hanoi - Vietnam - 2008
                Theoretical Problem No. 1

                    Figure 1
     A water-powered rice-pounding mortar

                                                               2

           39th International Physics Olympiad - Hanoi - Vietnam - 2008
                           Theoretical Problem No. 1

OPERATION CYCLE OF A WATER-POWERED RICE-POUNDING MORTAR

a)
                                           a) At the beginning there is no water in
                                           the bucket, the pestle rests on the mortar.
                                           Water flows into the bucket with a small
                                           rate, but for some time the lever remains
                                           in the horizontal position.

b)
                                           b) At some moment the amount of water
                                           is enough to lift the lever up. Due to the
                              α1
                                           tilt, water rushes to the farther side of the
                                           bucket, tilting the lever more quickly.

                                           Water starts to flow out at α = α1 .
c)

                                           c) As the angle α increases, water
                         α=β
                                           starts to flow out. At some particular tilt
                                           angle, α = β , the total torque is zero.

d)
                                           d)    α     continues     increasing,   water
                                           continues to flow out until no water
                        α2                 remains in the bucket.

                                           e) α      keeps increasing because of
                       α0                  inertia. Due to the shape of the bucket,
e)                                         water falls into the bucket but
                                           immediately flows out. The inertial
                                           motion of the lever continues until α

                                           reaches the maximal value α 0 .

                                           f) With no water in the bucket, the
f)
                                           weight of the lever pulls it back to the
                                           initial horizontal position. The pestle
         Figure 2                          gives the mortar (with rice inside) a
                                           pound and a new cycle begins.

                                                                                3

                       39th International Physics Olympiad - Hanoi - Vietnam - 2008
                                       Theoretical Problem No. 1

C. The problem
    Consider a water-powered rice-pounding mortar with the following parameters
(Figure 3)
        The mass of the lever (including the pestle but without water) is M = 30 kg,
        The center 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 , the center
of mass of the water body is denoted as N.
       The tilt angle of the lever with respect to the horizontal axis is α .
        The main length measurements of the mortar and the bucket are as in Figure 3.
    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.

                                                      T
                     Bucket             a =20cm

                                                                 Lever
                                                                             8 cm
                                                                                Pestle
                                 N
                     L = 74 cm
                                                      G

                          h= 12 cm
      γ =300                            b =15cm                               Mortar

               Figure 3   Design and dimensions of the rice-pounding mortar

1. The structure of the mortar
  At the beginning, the bucket is empty, and the lever 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 the distance from the center of mass G of the lever to the rotation
axis T. It is known that GT is horizontal when the bucket is empty.
    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 α 2 .

Determine α1 and α 2 .

   1.3. Let μ (α ) be the total torque (relative to the axis T) which comes from the

                                                                                4

                      39th International Physics Olympiad - Hanoi - Vietnam - 2008
                                      Theoretical Problem No. 1

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. Parameters of the working mode
   Let water flow into the bucket with a flow rate Φ which is constant and small. The
amount of water flowing into the bucket when the lever is in motion is negligible. In
this part, neglect the change of the moment of inertia during the working cycle.

   2.1. Sketch a graph of the torque μ as a function of the angle α , μ (α ) , during

one operation cycle. Write down explicitly the values of μ (α ) at angle α1, α2, and

α = 0.
   2.2. From the graph found in section 2.1., discuss and give the geometric

interpretation of the value of the total energy Wtotal produced by μ (α ) and the work

Wpounding that is transferred from the pestle to the rice.

   2.3. From the graph representing μ versus α , estimate α 0 and Wpounding (assume

the kinetic energy of water flowing into the bucket and out of the bucket is negligible.)
You may replace curve lines by zigzag lines, if it simplifies the calculation.

3. The rest mode
    Let water flow into the bucket with a constant rate Φ , but one cannot neglect the
amount of water flowing into the bucket during the motion of the lever.
    3.1. Assuming the bucket is always overflown with water,
        3.1.1. Sketch a graph of the torque μ as a function of the angle α in the
vicinity of α = β . 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 Δα when

α = β + Δα , and Δα 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 + Δα ( Δα is small). Show that the motion is,
with good accuracy, harmonic oscillation. Compute the period τ .

                                                                                5

                     39th International Physics Olympiad - Hanoi - Vietnam - 2008
                                     Theoretical Problem No. 1

  3.2. At a given Φ , the bucket is overflown with water at all times only if the lever
moves sufficiently slowly. There is an upper limit on the amplitude of harmonic

oscillation, which depends on Φ . Determine the minimal value Φ1 of Φ (in kg/s) so

that the lever can make a harmonic oscillator motion with amplitude 1o.
    3.3. Assume that Φ is sufficiently large so that during the free motion of the lever

when the tilting angle decreases from α 2 to α1 the bucket is always overflown with

water. However, if Φ 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 Φ 2 for the

rice-pounding mortar to not work.

                                                                                6

Solution:
39th International Physics Olympiad - Hanoi - Vietnam - 2008

                                  Theoretical Problem No. 1 /Solution

                                           Solution

1. The structure of the mortar
    1.1. Calculating the distance TG

   The volume of water in the bucket is V = 1000cm3 = 10−3 m3 . The length of the

bottom of the bucket is d = L − h tan 600 = ( 0.74 − 0.12 tan 600 ) m = 0.5322 m .

(as the initial data are given with two significant digits, we shall keep only two
significant digits in the final answer, but we keep more digits in the intermediate steps).
The height c of the water layer in the bucket is calculated from the formula:

                           c                (d 2 + 2 3V / b)1/ 2 − d
                V = bcd + b c tan 600 ⇒ c =
                           2                           3
Inserting numerical values for V , b and d , we find c = 0.01228 m .
When the lever lies horizontally, the distance, on the horizontal axis, between the rotation
                                                               d c
axis and the center of mass of water N, is TH ≈ a +             + tan 60o = 0.4714 m , and
                                                               2 4
TG = ( m / M ) TH = 0.01571m (see the figure below).

            R
                                         H                            T
            S           P
                                         N                   K

Answer: TG = 0.016 m .

   1.2. Calculating the values of α1 and α 2 .
  When the lever tilts with angle α1 , water level is at the edge of the bucket. At that

point the water volume is 10−3 m3 . Assume PQ < d . From geometry V = hb × PQ / 2 ,

from which PQ = 0.1111m . The assumption                      PQ < d is obviously satisfied
( d = 0.5322 m ).

   To compute the angle α1 , we note that tan α1 = h / QS=h /( PQ+ 3h ). From this

we find α1 = 20.6o .

                                                                                1

                         39th International Physics Olympiad - Hanoi - Vietnam - 2008

                                         Theoretical Problem No. 1 /Solution

When the tilt angle is 30o , the bucket is empty: α 2 = 30o .

                                                                                G
                                                                     h                    T

              R                  N

                                            I       Q
                                     P
                  S

  1.3. Determining the tilt angle β of the lever and the amount of water in the bucket
m when the total torque μ on the lever is equal to zero
  Denote     PQ = x (m) . The amount of water in the bucket is
              xhb
m = ρ water       = 9 x (kg) .
               2
    μ = 0 when the torque coming from the water in the bucket cancels out the torque
coming from the weight of the lever. The cross section of the water in the bucket is the
triangle PQR in the figure. The center of mass N of water is located at 2/3 of the meridian
RI, therefore NTG lies on a straight line. Then: mg × TN = Mg × TG or
           m × TN = M × TG = 30 × 0.1571 = 0.4714                (1)
Calculating TN from x then substitute (1) :
                       2         x                 x           x
           TN = L + a − ( h 3 + ) = 0.94 − 0.08 3 − = 0.8014 −
                       3         2                 3           3

which implies m × TN = 9 x (0.8014 − x / 3) = −3 x 2 + 7.213x                           (2)

So we find an equation for x :
        −3 x 2 + 7.213x = 0.4714                                     (3)
The solutions to (3) are x = 2.337 and x = 0.06723 . Since x has to be smaller than
0.5322, we have to take x = x0 = 0.06723 and m = 9 x0 = 0.6051kg .
                  h
    tan β =           = 0.4362 , or β = 23.57 o .
              x+h 3
Answer: m = 0.61kg and β = 23.6o .

2. Parameters of the working mode

                                                                                2

                         39th International Physics Olympiad - Hanoi - Vietnam - 2008

                                    Theoretical Problem No. 1 /Solution

   2.1.Graphs of μ (α ) , α (t ) , and μ (t ) during one operation cycle.
   Initially when there is no water in the bucket, α = 0 , μ has the largest magnitude
equal to gM × TG = 30 × 9.81× 0.01571 = 4.624 N ⋅ m . Our convention will be that
the sign of this torque is negative as it tends to decrease α .
   As water flows into the bucket, the torque coming from the water (which carries
positive sign) makes μ increase until μ is slightly positive, when the lever starts to
lift up. From that moment, by assumption, the amount of water in the bucket is constant.
     The lever tilts so the center of mass of water moves away from the rotation axis,
leading to an increase of μ , which reaches maximum when water is just about to

overflow the edge of the bucket. At this moment α = α1 = 20.6o .

   A simple calculation shows that
         SI = SP + PQ / 2 = 0.12 × 1.732 + 0.1111/ 2 = 0.2634 m .
                           2
         TN = 0.20 + 0.74 − SI = 0.7644 m .
                           3

         μmax = (1.0 × TN − 30 × TG ) g cos 20.6o

             = (1.0 × 0.7644 − 30 × 0.01571) × 9.81 × cos 20.6o = 2.690 N ⋅ m .

Therefore μ max = 2.7 N ⋅ m .
   As the bucket tilts further, the amount of water in the bucket decreases, and when
α = β , μ = 0 . Due to inertia, α keeps increasing and μ keeps decreasing. The

bucket       is        empty       when          α = 30o         ,        when          μ   equals

−30 × g × TG × cos 30o = −4.0 N ⋅ m . After that α keeps increasing due to inertia to

α 0 ( μ = − gM TG cos α 0 = −4.62 cos α 0 N ⋅ m ), then quickly decreases to 0
( μ = −4.62 N ⋅ m ).
   On this basis we can sketch the graphs of α (t ) , μ (t ) , and μ (α ) as in the figure
below

                                                                                3

                       39th International Physics Olympiad - Hanoi - Vietnam - 2008

                                      Theoretical Problem No. 1 /Solution

                             μ
                      2.7 N.m                          A

                                                           23.6o
                             O                                           E        α
                                                   o
                                                20.6 B             30o       α0

                 -4.6 cosα0 N.m                                          D
                  -4.0 N.m                                          C
                  -4.6 N.m F

   2.2. The infinitesimal work produced by the torque μ (α ) is dW = μ (α )dα . The
energy obtained by the lever during one cycle due to the action of μ (α ) is

W = v∫ μ (α )dα , which is the area limited by the line μ (α ) . Therefore Wtotal is equal

to the area enclosed by the curve (OABCDFO) on the graph μ (α ) .
   The work that the lever transfers to the mortar is the energy the lever receives as it

moves from the position α = α o to the horizontal position α = 0 . We have Wpounding

equals to the area of (OEDFO) on the graph                                        μ (α ) . It is equal to
gM × TG × sin α 0 = 4.6 sin α 0          (J).

   2.3. The magnitudes of α 0 can be estimated from the fact that at point D the energy
of the lever is zero. We have
                   area (OABO) = area (BEDCB)
Approximating OABO by a triangle, and BEDCB by a trapezoid, we obtain:
         23.6 × 2.7 × (1/ 2) = 4.0 × [(α 0 − 23.6) + (α 0 − 30)] × (1/ 2) ,

which implies α 0 = 34.7 o . From this we find

                                  0

                                  ∫ − Mg × TG × cos α dα = 4.62 × sin 34.7 = 2.63
                                                                                o
Wpounding = area (OEDFO) =
                                 34.76

                                                                                4

                          39th International Physics Olympiad - Hanoi - Vietnam - 2008

                                       Theoretical Problem No. 1 /Solution

Thus we find Wpounding ≈ 2.6 J.                                                     μ

3. The rest mode
    3.1.
         3.1.1. The bucket is always overflown
with water. The two branches of μ (α ) in the
                                                                              β          α
vicinity of α = β           corresponding to
increasing and decreasing α coincide with
each other.
    The graph implies that α = β is a stable
equilibrium of the mortar.
        3.1.2. Find the expression for the torque μ when the tilt angle is α = β + Δα
( Δα is small ).
    The mass of water in bucket when the lever tilts with angle α                        is

                                  ⎛ 1        1 ⎞
m = (1/ 2) ρ bh PQ , where PQ = h ⎜      −      0 ⎟
                                                    . A simple calculation shows that
                                  ⎝ tan α tan 30 ⎠
when α       increases from        β     to   β + Δα , the mass of water increases by

           bh 2 ρ            bh 2 ρ
Δm = −              Δ α ≈ −           Δα . The torque μ acting on the lever when the tilt
          2 sin 2 α         2 sin 2 β
is β + Δα equals the torque due to Δm .

   We have μ = Δm × g × TN × cos ( β + Δα ) . TN is found from the equilibrium

condition of the lever at tilting angle β :
   TN = M × TG / m = 30 × 0.01571/ 0.605 = 0.779 m .
We find at the end μ = − 47.2× Δα N ⋅ m ≈ −47 × Δα N ⋅ m .
         3.1.3. Equation of motion of the lever
           d 2α
    μ=I           where μ = − 47 ×Δα , α = β + Δα , and I is the sum of moments
            dt 2
of inertia of the lever and of the water in bucket relative to the axis T. Here I is not
constant the amount of water in the bucket depends on α . When Δα is small, one can
consider the amount and the shape of water in the bucket to be constant, so I is
approximatey a constant. Consider water in bucket as a material point with mass 0.6 kg, a

simple    calculation     gives    I = 12 + 0.6 × 0.782 = 12.36 ≈ 12.4 kg m 2 . We have

                        d 2 Δα
− 47 × Δα = 12.4 ×             . That is the equation for a harmonic oscillator with period
                         dt 2

                                                                                5

                        39th International Physics Olympiad - Hanoi - Vietnam - 2008

                                   Theoretical Problem No. 1 /Solution

         12.4
τ = 2π        = 3.227 . The answer is thereforeτ = 3.2s .
          47
  3.2. Harmonic oscillation of lever (around α = β ) when bucket is always overflown.
Assume the lever oscillate harmonically with amplitude Δα 0 around α = β . At time
t = 0 , Δα = 0 , the bucket is overflown. At time dt the tilt changes by dα . We are
interested in the case dα < 0 , i.e., the motion of lever is in the direction of decreasing
α , and one needs to add more water to overflow the bucket. The equation of motion is:
Δα = −Δα 0 sin( 2π t / τ ) , therefore d ( Δα ) = dα = −Δα 0 ( 2π / τ )cos( 2π t / τ )dt .

   For the bucket to be overflown, during this time the amount of water falling to the

                                  bh 2 ρ      2Δα 0π bh 2 ρ dt     ⎛ 2π t ⎞
bucket should be at least dm = −         dα =                  cos ⎜      ⎟            ;   dm is
                                 2 sin β        2τ sin β           ⎝ τ ⎠
                                      2               2

                              π bh 2 ρΔα 0
maximum at t = 0 ,      dm0 =              dt .
                               τ sin 2 β
   The amount of water falling to the bucket is related to flow rate Φ ; dm0 = Φdt ,

              π bh 2 ρΔα 0
therefore Φ =              .
               τ sin 2 β
   An overflown bucket is the necessary condition for harmonic oscillations of the lever,
therefore the condition for the lever to have harmonic oscillations with ampltude 1o or

2π/360 rad is Φ ≥ Φ1 with

                          π bh 2 ρ 2π
                    Φ1 =              = 0.2309 kg/s
                         360τ sin 2 β
   So Φ1 = 0.23kg/s .

   3.3 Determination of Φ 2
   If the bucket remains overflown when the tilt decreases to 20.6o , then the amount of
water in bucket should reach 1 kg at this time, and the lever oscillate harmonically with

amplitude equal 23.6o − 20.6o = 3o . The flow should exceed 3Φ1 , therefore

     Φ 2 = 3 × 0.23 ≈ 0.7 kg/s .
   This is the minimal flow rate for the rice-pounding mortar not to work.

                                                                                6

WATER-POWERED RICE-POUNDING MORTAR


[User]
Kord sooritas noor füüsik eksperimendi, et leida süstlakolvile mõjuvat
hõõrdejõudu. Ta tõmbas $V_{0}=\SI{10}{ml}$ mahuga süstlasse \SI{5,0}{ml} õhku ja sulges siis
süstla otsa sõrmega. Seejärel tõmbas ta süstla kolvi näiduni
$V_{1}=\SI{9,2}{ml}$ ja
lasi sellel seejärel aeglaselt tagasi liikuda. Kolb liikus, kuni näiduks jäi $V_{2}=\SI{5,8}{ml}$.
Mõõtmisel selgus, et süstlakolvi sisediameeter oli $d=\SI{9}{mm}$ ja kraadiklaas
näitas, et ruumis oli $t=\SI{27}{\degreeCelsius}$, õhu suhteline niiskus $R=30\%$ ja õhurõhk
$p_{0}=\SI{103,6}{kPa}$. Milline oli süstlakolvile mõjuv hõõrdejõud?\\
\textit{Märkus.} Kuna tegu on praktilise probleemiga, siis ei pruugi kõik
algandmed vajalikud olla.

Solution:
Kirjutame ideaalse gaasi olekuvõrrandi süstla jaoks vahetult peale sõrmega sulgemist:
\[p_{0}V=nRT.\]
Peale kolvi välja tõmbamist ja vabastamist:
\[p_{2}V_{2}=nRT \Rightarrow p_{2}=p_{0}\frac{V}{V_{2}}. \]
Kui kolb (ristlõikepindalaga $S=\frac{\pi d^{2}}{4}$)peale vabastamist seiskub, siis on kolvi hõõrdejõud $F_{h}$ tasakaalustanud rõhkude vahest tekitatud jõu:
\[F_{h}=S(p_{0}-p_{2})=Sp_{0} \left(1-\frac{V}{V_{2}}\right)=\frac{\pi
	d^{2}}{4}p_{0} \left(1-\frac{V}{V_{2}}\right) \approx \SI{3,6}{N} \]

Süstal

Kord sooritas noor füüsik eksperimendi, et leida süstlakolvile mõjuvat
hõõrdejõudu. Ta tõmbas $V_{0}=\SI{10}{ml}$ mahuga süstlasse \SI{5,0}{ml} õhku ja sulges siis
süstla otsa sõrmega. Seejärel tõmbas ta süstla kolvi näiduni
$V_{1}=\SI{9,2}{ml}$ ja
lasi sellel seejärel aeglaselt tagasi liikuda. Kolb liikus, kuni näiduks jäi $V_{2}=\SI{5,8}{ml}$.
Mõõtmisel selgus, et süstlakolvi sisediameeter oli $d=\SI{9}{mm}$ ja kraadiklaas
näitas, et ruumis oli $t=\SI{27}{\degreeCelsius}$, õhu suhteline niiskus $R=30\%$ ja õhurõhk
$p_{0}=\SI{103,6}{kPa}$. Milline oli süstlakolvile mõjuv hõõrdejõud?\\
\textit{Märkus.} Kuna tegu on praktilise probleemiga, siis ei pruugi kõik
algandmed vajalikud olla.

Kolvi lõppasendis tasakaalustab kolvile mõjuv hõõrdejõud rõhkude vahest tekitatud rõhumisjõu.


[User]
01h   USAPhO+ 2021, problem 2. A full analysis of a real dark matter detector.

      Remark
      Problem 26 estimates the size of a star using a rough treatment of the nuclear physics
      we know applies at its core. When you think about it, it’s amazing that this is possible
      at all. There is essentially no way to directly probe what’s going on inside any star. We
      can basically only measure the size of the star, its rough age, and the temperature of its
      surface. But the models work! This was the result of generations of painstaking effort,
      which stimulated progress in both nuclear and particle physics. To learn the history of solar
      modeling, see this article for the early days and this article for the eventual triumph. For
      more estimates of the sizes of astronomical objects, see this article.

      It is even more remarkable when you realize that this sort of story applies to essentially all
      of the dozens of known types of astronomical objects. For most of these, astronomers can
      directly measure only the electromagnetic spectrum and its variation over time, but that little
      is enough to develop and test detailed physical models. Most people think astronomy is very
      concrete, because they constantly see stars in the night sky, and flashy “artist’s impression”
      graphics on the internet. But the field is actually extremely theoretical, with every insight
      won through extensive calculation.

Solution:
01h   USAPhO+ 2021, problem 2. A full analysis of a real dark matter detector.

      Remark
      Problem 26 estimates the size of a star using a rough treatment of the nuclear physics
      we know applies at its core. When you think about it, it’s amazing that this is possible
      at all. There is essentially no way to directly probe what’s going on inside any star. We
      can basically only measure the size of the star, its rough age, and the temperature of its
      surface. But the models work! This was the result of generations of painstaking effort,
      which stimulated progress in both nuclear and particle physics. To learn the history of solar
      modeling, see this article for the early days and this article for the eventual triumph. For
      more estimates of the sizes of astronomical objects, see this article.

      It is even more remarkable when you realize that this sort of story applies to essentially all
      of the dozens of known types of astronomical objects. For most of these, astronomers can
      directly measure only the electromagnetic spectrum and its variation over time, but that little
      is enough to develop and test detailed physical models. Most people think astronomy is very
      concrete, because they constantly see stars in the night sky, and flashy “artist’s impression”
      graphics on the internet. But the field is actually extremely theoretical, with every insight
      won through extensive calculation.


[User]
Rong sõidab kiirusega $v = \SI{100}{km/h}$ ja pidurdab järsult (blokeerides rattad). Graafikul on toodud rongi rataste ja rööbaste vahelise hõõrdeteguri $\mu$ sõltuvus kiirusest (\si{km/h}).\\
\osa Kui pikk on rongi täieliku peatumiseni kulunud aeg?\\
\osa Kui suur on pidurdusmaa pikkus? Mõlemad vastused tuleb leida
graafikualuste pindaladena sobilikult valitud teljestikes.

\begin{center}
	\includegraphics[width=\linewidth]{2006-lahg-10-yl}
\end{center}

Solution:
\osa Vaatame väikest kiiruste vahemikku $\Delta v$, mille sees võib kiirust $v$ ja järelikult ka hõõrdetegurit $\mu (v)$ lugeda konstantseks. Kiirendus on selle liikumisfaasi jooksul siis $a = \mu (v)g$ ning kiiruse vähenemiseks kuuluv aeg
\[
\Delta t=\frac{\Delta v}{a}=\frac{\Delta v}{\mu(v) g}.
\]

\begin{center}
	\includegraphics[width=\linewidth]{2006-lahg-10-lah1}
\end{center}

Summaarne aeg oleks summa üle kõigi selliste väikeste kiirusemuutuste. Konstrueerime esialgsest graafikust lähtudes suuruse $1/\mu (v)$ sõltuvuse kiirusest $v$ (vt joonist). Siis pidurdamiseks kuuluv aeg on selle sõltuvuse graafikualune pindala (kiirusest \num{0} kuni \SI{100}{km/h}), jagatud $g$-ga. Seejuures peame silmas, et kiirus $x$-teljel peab olema meetrites sekundis (\si{m/s}). Vastuseks saame ligikaudselt \SI{18}{s}.

\osa Vaatame samasugust kiiruste vahemikku $\Delta v$ nagu esimeses osas. Ajaga $\Delta t$, mille jooksul kiirus selle võrra väheneb, läbib rong teepikkuse
\[
\Delta s = v\Delta t = \frac{v\Delta v}{\mu (v)g}.
\]
Summaarne pidurdusmaa oleks summa üle kõigi selliste väikeste teepikkuste. Konstrueerime esialgsest graafikust lähtudes suuruse $v/\mu (v)$ sõltuvuse kiirusest $v$ (vt joonist). Siis pidurdusmaa on selle sõltuvuse graafikualune pindala (kiirusest \num{0} kuni \SI{100}{km/h}), jagatud $g$-ga. Kiirus $x$-teljel on samuti meetrites sekundis (\si{m/s}). Vastuseks saame ligikaudselt \SI{235}{m}.

\begin{center}
	\includegraphics[width=\linewidth]{2006-lahg-10-lah2}
\end{center}

\emph{Märkus}. Integraalne avaldis pidurdusaja jaoks on kujul
\[
t=\int_{0}^{v_{0}} \frac{\D v}{\mu(v) g}=\frac{1}{g} \int_{0}^{v_{0}} \frac{\D v}{\mu(v)},
\]
kus $v_0$ on rongi algkiirus. Avaldis pidurdusmaa pikkuse jaoks on aga
\[
s=\int_{0}^{v_{0}} \frac{v \D v}{\mu(v) g}=\frac{1}{g} \int_{0}^{v_{0}} \frac{v \D v}{\mu(v)}.
\]

Rong

Rong sõidab kiirusega $v = \SI{100}{km/h}$ ja pidurdab järsult (blokeerides rattad). Graafikul on toodud rongi rataste ja rööbaste vahelise hõõrdeteguri $\mu$ sõltuvus kiirusest (\si{km/h}).\\
\osa Kui pikk on rongi täieliku peatumiseni kulunud aeg?\\
\osa Kui suur on pidurdusmaa pikkus? Mõlemad vastused tuleb leida
graafikualuste pindaladena sobilikult valitud teljestikes.

\begin{center}
	\includegraphics[width=\linewidth]{2006-lahg-10-yl}
\end{center}

Otsitavate suuruste jaoks on kasulik vaadelda lühikest ajavahemikku $\Delta t$, mille jooksul on rongi kiirus, ja seega hõõrdetegur, ligikaudu konstantsed. Seejärel saab saadud ajavahemikke summeerida terve graafiku ulatuses. Selle jaoks peab vajadusel konstrueerima uued graafikud teistsuguste telgedega.


[User]
USAPhO 2012, problem B2.

       Idea 2
       Not all motional emfs can be found using E = −dΦ/dt. Sometimes, for more complex
       geometries where there is no clear “loop”, it’s easier to go back to the Lorentz force law.

Solution:
USAPhO 2012, problem B2.
       Idea 2
       Not all motional emfs can be found using E = −dΦ/dt. Sometimes, for more complex
       geometries where there is no clear “loop”, it’s easier to go back to the Lorentz force law.


[User]
4. Questions consist of sub-questions. Write your detailed answer in the blank space
   provided below the sub-question and final answer to the sub-question in the smaller
   box which follows the blank space.

Solution:
4. (a) Taking q = q = e, rn = n a − β
                     1           2                            0

                                q1 q2
      (b) En =                       2
              − 8πǫ (n a − β )
                            0            0

      (c) ∆E =11.5 eV (±0.15 eV acceptable)

         dv             
                   + v B
                             mv           

INPhO 2011 Problem 4


[User]
A 2x2 grid is made out of wire (see figure), the resistance of each small square’s edge is $r=\SI{1}{\ohm}$. Find the resistance between the points A and B.
\begin{center}
\includegraphics[width=0.2\linewidth]{2014-v3g-01-ruudustik}
\end{center}

Solution: Sümmeetria tõttu on ruudustiku kaks tähistamata nurka ja keskpunkt sama potentsiaaliga, mistõttu võid need kolm punkti kokku ühendada. Saadud skeem jaotub ilusti rööp- ja jadaühendusteks, nende abil saame punktide A ja B vaheliseks takistuseks $R=\SI{1,5}{\ohm}$.

Ruudustik

Traadist on valmistatud 2x2 ruudustik (vt joonist), iga väikese ruudu külje takistus on $r=\SI{1}{\ohm}$. Leidke punktide A ja B vaheline takistus.

\begin{center}
\includegraphics[width=0.2\linewidth]{2014-v3g-01-ruudustik}
\end{center}

Sümmeetria kaalutlustel saab skeemis sama potentsiaaliga punkte kokku ühendada.