[User]
01^   IPhO 2014, problem 1B.

       Idea 2
       One can also have liquid, solid, and air in the same problem, which leads to some complications.
       Let Al and As be the surface areas of the liquid and solid exposed to the air, and Asl be the
       surface area of the liquid-solid interface. Then there are three terms in the work,

                                      d̄W = γl dAl + γs dAs + γsl dAsl .

       In other words, there are three surface tensions, one associated with each kind of interface.

       Both γl and γs arise from the fact that cohesive liquid-liquid or solid-solid bonds are broken
       to form a surface. However, γsl is determined by the adhesive forces between the liquid and
       solid, which may lead to a positive or negative contribution to the energy.

       Specifically, let’s define the energy of adhesion Usl to be the work needed, per area, to separate
       a liquid from a solid, thereby turning a liquid-solid interface into a liquid-air and solid-air
       interface. By the definitions above,

                                             Usl = γl + γs − γsl .

       Now, Usl can be computed in terms of microscopic chemical bond energies, like γl and γs ,
       so this result can also be thought of as an microscopic definition of γsl . When a liquid is in
       contact with a solid, the solid exerts a force per length of Usl on the boundary of the liquid,
       along the solid.

       Example 1

       The surface of a drop of water makes a contact angle θ with a solid, as shown.

       When θ is acute, the surface is said to be hydrophilic. If θ is obtuse, it is hydrophobic. Find
       an expression for θ in terms of the relevant surface tensions.

  Solution
  If the liquid drop expands outward by δx, the areas of various surfaces change, as shown.

  The change in energy is
                                 dU ∝ γsl δx + γl cos θ δx − γs δx
  and this must be equal to zero in equilibrium. Thus,
                                              γs − γsl   Usl
                                    cos θ =            =     − 1.
                                                 γl      γl
  This is Young’s equation. The liquid surface tension γl must be positive; otherwise the
  liquid could not exist stably at all, but rather would disperse into gas. Thus, the surface is
  hydrophilic when Usl > γl and hydrophobic when Usl < γl .

  As extreme cases, note that there is no solution for θ when Usl > 2γl . In this limit, the
  surface is so hydrophilic that the liquid spreads out and coats the entire solid; this is
  known as perfect wetting. There is also no solution when Usl < 0, in which case the liquid
  disperses into many tiny nearly spherical drops, each with a tiny area of contact with the solid.

  This derivation was in terms of energy, which is typically easier for surface tension. The
  same result can be derived in terms of forces, but it’s more subtle than it looks; the standard
  derivation in textbooks is wrong. For a clear derivation, see section 9.3 of Wang and Ricardo.

  Example 2

  A very thin, hollow glass tube of radius r is dipped vertically inside a container of water.

  Find the equilibrium height of the water in the tube.

  Solution
  We first encountered this problem in M7, where we solved it by using Pascal’s principle,
  giving an answer in terms of the contact angle. The derivation above of the contact angle
  completes this solution. However, we can also solve the problem using energy or force.

  In terms of energy, if we move the height of the water up by δh, then

                             dU = ρπr2 gh δh + (γsl − γs )2πr δh = 0

  and solving gives
                                        2(γs − γsl )   2γl cos θ
                                   h=                =
                                           ρgr           ρgr
  using Young’s equation. Alternatively, in terms of force, consider the vertical forces acting
  on the column of water inside the tube. There is an upward force of adhesion from the solid
  wall of 2πrUsl , and a downward surface tension force from the liquid below of 2πrγl . Then

                                 F = 2πr(Usl − γl ) − ρπr2 gh = 0

  which yields precisely the same result.

  Example 3

  Fill a dish with water, and sprinkle something small over it, such as ground pepper. If you
  place a drop of detergent in the middle of the dish, then the pepper will “flee” away to the
  edges. Why does this happen?

  Solution
  Detergent is a surfactant, meaning that it decreases the surface tension of water. When one
  places the detergent in the middle of the dish, it diffuses outward, making the surface tension
  temporarily higher near edges of the dish. This leads to an unbalanced surface tension force
  on the pepper grains, pulling them to the edges.

  This phenomenon is called the Marangoni effect. Of course, the force vanishes once the
  detergent becomes uniform distributed, and the surface tension is uniform again.

  Remark
  Here’s a neat fact: the number of atoms that fit into a drop of water is comparable to the
  number of drops of water that fit inside the tallest mountains. We can show this using rough
  estimates, in the style of P1. Let Eb be the energy of a typical chemical bond, let m be the
  mass of an atom, and let d be the typical distance between atoms.

  The size ℓ of a droplet of water, such as one that drips from a leaky ceiling, is the size where
  surface tension forces balance gravitational ones. By dimensional analysis, we must have
                                                p
                                            ℓ ∼ γ/ρg

    as we showed in M7. Now, ρ ∼ m/d3 , and the logic of problem 1 implies γ ∼ Eb /d2 , so
                                           p
                                       ℓ ∼ Eb d/mg.

    Now consider the height H of the tallest mountains. The height of mountains is limited by
    the rigidity of rock; if the pressure is too great, then the rock underneath the mountain will
    deform, causing it to sink into the ground. Let’s consider an atom-thick column of this rock.
    If it sunk down by a distance d, then the gravitational potential energy harvested would be
    mgH. However, the atom at the bottom would have to break its chemical bonds with its
    horizontal neighbors, which takes energy Eb . Balancing these gives a maximum height

                                             H ∼ Eb /mg.

    We have therefore shown that                    √
                                               ℓ∼       Hd
    which implies the original statement, within a few orders of magnitude.

2    Melting, Freezing, Boiling, Evaporation, and Condensation
    Idea 3
    A phase transition is a sudden, dramatic change in a system as thermodynamic variables
    such as the temperature are varied. Most of the ones you’ll see have a latent heat

                                               Q = mL.

    For example, if ice is heated up, its temperature will gradually increase until it hits 0 ◦ C. At
    that point, the temperature will remain constant until all of the ice is melted, i.e. when the
    full latent heat has been supplied.

    Remark
    We can roughly estimate the latent heats of melting and evaporation. In general, the latent
    heat can go into either in breaking molecular bonds, or increasing the entropy.

    When a solid melts into a liquid, the molecules stay right next to each other, so changing
    bond energy isn’t the dominant effect. Instead, it’s the increase in entropy as the liquid
    molecules become free to rotate. Let’s suppose that the molecules each gain a few extra
    possible quantum states. This corresponds to an entropy increase per molecule ∆S ∼ kB ,
    which means a latent heat per mass of
                                            T ∆S   kB T   RT
                                      L=         ∼      =
                                            mmol   mmol    µ
    where µ is the molar mass, or equivalently a latent heat per mole L ∼ RT . For water, we
    get L ∼ 2 × 105 J/kg, which is of the same order of magnitude as the true value 3.3 × 105 J/kg.

       When a liquid becomes a gas, the dominant effect is typically the huge increase in entropy
       kB log(Vgas /Vliq ) per molecule because they get much more space to move. The ratio inside
       the logarithm is huge, which means that while the volumes per molecule Vgas and Vliq vary by
       order-one amounts between phase transitions, the logarithm of their ratio is always around
       the same value, which turns out to be about 10. This gives a latent heat per mass of
                                                10kB T   10RT
                                           L∼          =      .
                                                 mmol      µ
       This result is called Trouton’s rule, and it is surprisingly accurate for most liquids. However,
       the latent heat of vaporization for water is noticeably higher, L = 2.26 × 106 J/kg. This is
       because of the extra energy needed to break hydrogen bonds.

Solution:
01^   IPhO 2014, problem 1B.

       Idea 2
       One can also have liquid, solid, and air in the same problem, which leads to some complications.
       Let Al and As be the surface areas of the liquid and solid exposed to the air, and Asl be the
       surface area of the liquid-solid interface. Then there are three terms in the work,

                                      d̄W = γl dAl + γs dAs + γsl dAsl .

       In other words, there are three surface tensions, one associated with each kind of interface.

       Both γl and γs arise from the fact that cohesive liquid-liquid or solid-solid bonds are broken
       to form a surface. However, γsl is determined by the adhesive forces between the liquid and
       solid, which may lead to a positive or negative contribution to the energy.

       Specifically, let’s define the energy of adhesion Usl to be the work needed, per area, to separate
       a liquid from a solid, thereby turning a liquid-solid interface into a liquid-air and solid-air
       interface. By the definitions above,

                                             Usl = γl + γs − γsl .

       Now, Usl can be computed in terms of microscopic chemical bond energies, like γl and γs ,
       so this result can also be thought of as an microscopic definition of γsl . When a liquid is in
       contact with a solid, the solid exerts a force per length of Usl on the boundary of the liquid,
       along the solid.

       Example 1

       The surface of a drop of water makes a contact angle θ with a solid, as shown.

  When θ is acute, the surface is said to be hydrophilic. If θ is obtuse, it is hydrophobic. Find
  an expression for θ in terms of the relevant surface tensions.

  Solution
  If the liquid drop expands outward by δx, the areas of various surfaces change, as shown.

  The change in energy is
                                 dU ∝ γsl δx + γl cos θ δx − γs δx
  and this must be equal to zero in equilibrium. Thus,
                                              γs − γsl   Usl
                                    cos θ =            =     − 1.
                                                 γl      γl
  This is Young’s equation. The liquid surface tension γl must be positive; otherwise the
  liquid could not exist stably at all, but rather would disperse into gas. Thus, the surface is
  hydrophilic when Usl > γl and hydrophobic when Usl < γl .

  As extreme cases, note that there is no solution for θ when Usl > 2γl . In this limit, the
  surface is so hydrophilic that the liquid spreads out and coats the entire solid; this is
  known as perfect wetting. There is also no solution when Usl < 0, in which case the liquid
  disperses into many tiny nearly spherical drops, each with a tiny area of contact with the solid.

  This derivation was in terms of energy, which is typically easier for surface tension. The
  same result can be derived in terms of forces, but it’s more subtle than it looks; the standard
  derivation in textbooks is wrong. For a clear derivation, see section 9.3 of Wang and Ricardo.

  Example 2

  A very thin, hollow glass tube of radius r is dipped vertically inside a container of water.

  Find the equilibrium height of the water in the tube.

  Solution
  We first encountered this problem in M7, where we solved it by using Pascal’s principle,
  giving an answer in terms of the contact angle. The derivation above of the contact angle
  completes this solution. However, we can also solve the problem using energy or force.

  In terms of energy, if we move the height of the water up by δh, then

                             dU = ρπr2 gh δh + (γsl − γs )2πr δh = 0

  and solving gives
                                        2(γs − γsl )   2γl cos θ
                                   h=                =
                                           ρgr           ρgr
  using Young’s equation. Alternatively, in terms of force, consider the vertical forces acting
  on the column of water inside the tube. There is an upward force of adhesion from the solid
  wall of 2πrUsl , and a downward surface tension force from the liquid below of 2πrγl . Then

                                F = 2πr(Usl − γl ) − ρπr2 gh = 0

  which yields precisely the same result.

  Example 3

  Fill a dish with water, and sprinkle something small over it, such as ground pepper. If you
  place a drop of detergent in the middle of the dish, then the pepper will “flee” away to the
  edges. Why does this happen?

  Solution
  Detergent is a surfactant, meaning that it decreases the surface tension of water. When one
  places the detergent in the middle of the dish, it diffuses outward, making the surface tension
  temporarily higher near edges of the dish. This leads to an unbalanced surface tension force
  on the pepper grains, pulling them to the edges.

    This phenomenon is called the Marangoni effect. Of course, the force vanishes once the
    detergent becomes uniform distributed, and the surface tension is uniform again.

    Remark
    Here’s a neat fact: the number of atoms that fit into a drop of water is comparable to the
    number of drops of water that fit inside the tallest mountains. We can show this using rough
    estimates, in the style of P1. Let Eb be the energy of a typical chemical bond, let m be the
    mass of an atom, and let d be the typical distance between atoms.

    The size ℓ of a droplet of water, such as one that drips from a leaky ceiling, is the size where
    surface tension forces balance gravitational ones. By dimensional analysis, we must have
                                                  p
                                              ℓ ∼ γ/ρg

    as we showed in M7. Now, ρ ∼ m/d3 , and the logic of problem 1 implies γ ∼ Eb /d2 , so
                                           p
                                       ℓ ∼ Eb d/mg.

    Now consider the height H of the tallest mountains. The height of mountains is limited by
    the rigidity of rock; if the pressure is too great, then the rock underneath the mountain will
    deform, causing it to sink into the ground. Let’s consider an atom-thick column of this rock.
    If it sunk down by a distance d, then the gravitational potential energy harvested would be
    mgH. However, the atom at the bottom would have to break its chemical bonds with its
    horizontal neighbors, which takes energy Eb . Balancing these gives a maximum height

                                             H ∼ Eb /mg.

    We have therefore shown that                    √
                                               ℓ∼       Hd
    which implies the original statement, within a few orders of magnitude.

2    Melting, Freezing, Boiling, Evaporation, and Condensation
    Idea 3
    A phase transition is a sudden, dramatic change in a system as thermodynamic variables
    such as the temperature are varied. Most of the ones you’ll see have a latent heat

                                               Q = mL.

    For example, if ice is heated up, its temperature will gradually increase until it hits 0 ◦ C. At
    that point, the temperature will remain constant until all of the ice is melted, i.e. when the
    full latent heat has been supplied.

       Remark
       We can roughly estimate the latent heats of melting and evaporation. In general, the latent
       heat can go into either in breaking molecular bonds, or increasing the entropy.

       When a solid melts into a liquid, the molecules stay right next to each other, so changing
       bond energy isn’t the dominant effect. Instead, it’s the increase in entropy as the liquid
       molecules become free to rotate. Let’s suppose that the molecules each gain a few extra
       possible quantum states. This corresponds to an entropy increase per molecule ∆S ∼ kB ,
       which means a latent heat per mass of
                                              T ∆S   kB T   RT
                                         L=        ∼      =
                                              mmol   mmol    µ
       where µ is the molar mass, or equivalently a latent heat per mole L ∼ RT . For water, we
       get L ∼ 2 × 105 J/kg, which is of the same order of magnitude as the true value 3.3 × 105 J/kg.

       When a liquid becomes a gas, the dominant effect is typically the huge increase in entropy
       kB log(Vgas /Vliq ) per molecule because they get much more space to move. The ratio inside
       the logarithm is huge, which means that while the volumes per molecule Vgas and Vliq vary by
       order-one amounts between phase transitions, the logarithm of their ratio is always around
       the same value, which turns out to be about 10. This gives a latent heat per mass of
                                                10kB T   10RT
                                           L∼          =      .
                                                 mmol      µ
       This result is called Trouton’s rule, and it is surprisingly accurate for most liquids. However,
       the latent heat of vaporization for water is noticeably higher, L = 2.26 × 106 J/kg. This is
       because of the extra energy needed to break hydrogen bonds.


[User]
Question B1
The Muon Shot
In 2023, American particle physicists recommended developing a muon collider to investigate the
nature of fundamental particles. Such a collider requires much less space then other options because
of the muon’s high mass m, which makes it easier to accelerate to a very high energy E        mc2 .

a. When a muon colides head-on with an antimuon, which has the same energy and mass, a new
   particle of mass 2E/c2 can be produced. If the antimuon was instead at rest, what energy would
   the muon need to produce such a particle?

Unfortunately, muons and antimuons are unstable, with lifetime ⌧ . That is, if one such particle
exists at time t = 0, then in its rest frame, the probability it has not decayed by time t is e t/⌧ .

b. Suppose the muons begin at rest, and are accelerated so that each muon’s energy increases at a
   very large, constant rate ↵ in the lab frame. Find the fraction f of muons that have decayed by
   the time each muon has energy E, assuming f is small.

The collider produces a “bunch” of muons with energy E, uniformly distributed in a thin disc
of radius R = 10 6 m. It also simultaneously produces a similar “antibunch” of antimuons. For
simplicity, model each muon and antimuon as a sphere of radius r = 10 21 m, and suppose a
muon-antimuon collision occurs whenever two such spheres touch.

c. Initially, the bunch and antibunch each contain N = 1014 particles. If they immediately collide
   head-on, what is the average number of muon-antimuon collisions, to one significant figure?

d. The bunch travels clockwise along a ring of circumference ` = 10 km, while the antibunch travels
   along the same path in the opposite direction. Assume all particles maintain a constant energy
   E = 105 mc2 , and that the muon lifetime is ⌧ = 2.2 ⇥ 10 6 s. To one significant figure, what is
   the average number of muon-antimuon collisions that occur before all of the particles decay?

                   Copyright ©2024 American Association of Physics Teachers

2024 USAPhO                                    Part B                                                9

Solution:
Question B1
The Muon Shot
In 2023, American particle physicists recommended developing a muon collider to investigate the
nature of fundamental particles. Such a collider requires much less space then other options because
of the muon’s high mass m, which makes it easier to accelerate to a very high energy E ≫ mc2 .

a. When a muon colides head-on with an antimuon, which has the same energy and mass, a new
   particle of mass 2E/c2 can be produced. If the antimuon was instead at rest, what energy would
   the muon need to produce such a particle?

                                               Solution
     The shortest solution involves setting c = 1 and using four-vectors. In the original situation,
     the muon and antimuon have four-momenta pµ1 and pµ2 , where

                                  p1 · p2 = (E, p) · (E, −p) = E 2 + p2 .

     To have an equivalent collision where the antimuon is at rest, we boost this configuration to
     the antimuon’s rest frame, where p′µ       ′ ′        ′µ
                                         1 = (E , p ) and p2 = (m, 0). The inner product of the
     four-vectors stays the same, so

                                         p′1 · p′2 = E ′ m = p1 · p2 .

     Since E ≫ m, we have p2 ≈ E 2 , so solving for E ′ and restoring the factors of c gives

                                                        2E 2
                                                E′ =         .
                                                        mc2
     This is much greater than E, so it is practical to accelerate the antimuon as well. Incidentally,
     you can also solve the problem exactly, in which case you’ll get

                                               2E 2         2E 2
                                        E′ =        − mc2
                                                          ≈      .
                                               mc2          mc2
     Either answer is acceptable.

Unfortunately, muons and antimuons are unstable, with lifetime τ . That is, if one such particle
exists at time t = 0, then in its rest frame, the probability it has not decayed by time t is e−t/τ .

b. Suppose the muons begin at rest, and are accelerated so that each muon’s energy increases at a
   very large, constant rate α in the lab frame. Find the fraction f of muons that have decayed by
   the time each muon has energy E, assuming f is small.

                                               Solution
     Continuing to set c = 1, the energy of the muon in the lab frame is E(t) = αt + m, but we

                   Copyright ©2024 American Association of Physics Teachers

2024 USAPhO                                     Part B                                            14

     also know that E(t) = γ(t)m, so that
                                                         αt
                                               γ =1+        .
                                                         m
     The acceleration process begins at time t = 0, and ends at time tf = (E − m)/α. Accounting
     for time dilation, the amount of time that elapses in the muon’s frame is
                                   Z tf
                                              m tf
                                                Z
                                        dt               1        m E
                              τf =         =                  =     ln .
                                    0   γ     α 0 t + m/α         α    m

     To relate this to f , we note that
                                                                τf
                                           f = 1 − e−τf /τ ≈
                                                                τ
     where the second step uses the assumption that f is small. Therefore,

                                                  mc2     E
                                             f=       ln
                                                  ατ     mc2
     where we restored the factors of c.

The collider produces a “bunch” of muons with energy E, uniformly distributed in a thin disc
of radius R = 10−6 m. It also simultaneously produces a similar “antibunch” of antimuons. For
simplicity, model each muon and antimuon as a sphere of radius r = 10−21 m, and suppose a
muon-antimuon collision occurs whenever two such spheres touch.

c. Initially, the bunch and antibunch each contain N = 1014 particles. If they immediately collide
   head-on, what is the average number of muon-antimuon collisions, to one significant figure?

                                               Solution
     Consider one muon and antimuon. A collision occurs when their centers are separated by
     less than 2r. Fixing the location of the muon, the probability that the antimuon is within
     the appropriate area is approximately π(2r)2 /(πR2 ) = (2r/R)2 , since r ≪ R. Each pair of
     muons and antimuons has the same chance to collide, so the expected number of collision
     events is
                                             2rN 2

                                                    = 0.04.
                                              R
     This is much smaller than N , which justifies our assumption that the collision events are
     independent. It might seem odd for the answer to be less than 1, but this is desired, as
     having many collisions occur at once would make it hard to see what happens in each
     collision.

d. The bunch travels clockwise along a ring of circumference ℓ = 10 km, while the antibunch travels
   along the same path in the opposite direction. Assume all particles maintain a constant energy
   E = 105 mc2 , and that the muon lifetime is τ = 2.2 × 10−6 s. To one significant figure, what is
   the average number of muon-antimuon collisions that occur before all of the particles decay?

                   Copyright ©2024 American Association of Physics Teachers

2024 USAPhO                                        Part B                                      15

                                               Solution
    If the number of particles remaining in the bunch and the antibunch is Nk , where k = 0 for
    the first collision, then the expected number of collisions is
                                                     ∞
                                                   2 X          2
                                             2rN            Nk
                                                                      .
                                              R             N
                                                      k=0

   We found above that only a small number of collisions occurs per bunch-antibunch crossing,
   so the decrease in Nk over time is almost entirely due to decay. A crossing occurs every time
   each bunch or antibunch traverses half of the ring, corresponding to a proper time increment

                                             ℓ/(2c)
                                   ∆τ =             = 1.67 × 10−10 s.
                                              105
   Therefore, since τ ≫ ∆τ , we have
                   ∞              ∞
                       Nk 2

                   X               X                          1                 τ
                               =         e−2k ∆τ /τ =                     ≈        = 6600.
                   k=0
                         N
                                   k=0
                                                        1 − e−2 ∆τ /τ         2 ∆τ

   The expected total number of collisions is 260, which rounds to 300.
   The rough numbers given here correspond to a muon collider which would be able to probe
   new particles 10 times as heavy as those probed at the existing Large Hadron Collider.
   Roughly one in a million muon-antimuon collisions yield a Higgs boson, so that an enormous
   number of them can be produced for detailed study. Remarkably, such a muon collider could
   be smaller in size than the LHC, while other proposals involving electrons or protons would
   need to be about 10 times longer. However, given the muon’s short lifetime, it may be very
   hard to create the required focused muon beams. The feasibility of this “muon shot” is
   currently being investigated by particle physicists around the world.
    One of us (TB) thanks Nathaniel Craig, Andrew Fee and Sergo Jindariani for discussions
    during our work on this problem.

                 Copyright ©2024 American Association of Physics Teachers

2024 USAPhO                                    Part B                                               16

USAPhO 2024 Question B1


[User] Kolm ratturit sõitsid linnast $A$ linna $B$. Linnast $A$ väljusid nad üheaegselt. Esimese ratturi keskmine kiirus oli $v_1 = \SI{30}{km/h}$, teise ratturi oma $v_2 = \SI{20}{km/h}$. Esimene rattur jõudis sihtpunkti kell 19.00, teine rattur kell 20.00 ning kolmas rattur kell 21.00. Milline oli kolmanda ratturi keskmine kiirus $v_3$?

Solution:
Olgu esimese ratturi sõiduaeg tundides $t$, siis teise ratturi sõiduaeg on $t + 1$ ning kolmanda ratturi oma $t+2$. Et esimene ja teine rattur sõitsid läbi sama pika vahemaa, saame võrrandi
\[
30 t=20(t+1),
\]
kust saame $t = \num{2}$ tundi. Järelikult linnade $A$ ja $B$ vaheline kaugus on $s = \num{30}·\num{2} = \SI{60}{km}$ kilomeetrit. Kolmanda ratturi keskmine kiirus oli seega
\[
v_{3}=\frac{s}{t+2}=\SI{15}{km/h}.
\]

Ratturid

Kolm ratturit sõitsid linnast $A$ linna $B$. Linnast $A$ väljusid nad üheaegselt. Esimese ratturi keskmine kiirus oli $v_1 = \SI{30}{km/h}$, teise ratturi oma $v_2 = \SI{20}{km/h}$. Esimene rattur jõudis sihtpunkti kell 19.00, teine rattur kell 20.00 ning kolmas rattur kell 21.00. Milline oli kolmanda ratturi keskmine kiirus $v_3$?

Ratturite sõiduaeg on leitav asjaolust, et esimene ja teine rattur läbisid sama pika vahemaa.


[User]
1. The Flywheel Chronicles
   In the following experiment we are interested in determining the moment of inertia of a ﬂywheel.
   The free ends of the axle in a ﬂywheel assembly are placed inside grooves at both ends, to rigid
   supports provided on the wall (see diagram below). The diameter of the axle is d = 2.72 cm. The
   total work done by the axle in overcoming the friction in the two grooves per rotation is W . A
   massless string, attached to a point mass m is wound tightly in n turns (without overlap) on the
   axle. The string unwinds from the axle without slipping as the mass descends from an initial
   height h. The length of the string is adjusted such that when the mass just touches the ﬂoor, the
   string detaches from the axle.

                                                          Flywheel

                              Axle

                                                                         d

                                     m
                                                                                Rigid support
              Rigid support

                                                  h
                                                                E
                                           CS
                  HB
  From the instant when the mass touches the ﬂoor (taken as t = 0), the ﬂywheel continues to
  rotate, adding another N number of rotations before coming to rest in time t = T . The ﬁgure is
  not to scale.
   (a) [5 marks] Derive an expression for the moment of inertia I of the ﬂywheel in terms of
       m, N, T , and other known parameters.
   (b) [3 marks] The following data has been obtained in this experiment.

                                         m (gm)       N        T (sec)

                                         150.0        145.25   190.0

                                         200.0        200.00   225.5

                                         250.0        238.50   235.5

         Calculate the value of I for n = 16, and h = 139 cm.

Solution:
1. The Flywheel Chronicles
   In the following experiment we are interested in determining the moment of inertia of a ﬂywheel.
   The free ends of the axle in a ﬂywheel assembly are placed inside grooves at both ends, to rigid
   supports provided on the wall (see diagram below). The diameter of the axle is d = 2.72 cm. The
   total work done by the axle in overcoming the friction in the two grooves per rotation is W . A
   massless string, attached to a point mass m is wound tightly in n turns (without overlap) on the
   axle. The string unwinds from the axle without slipping as the mass descends from an initial
   height h. The length of the string is adjusted such that when the mass just touches the ﬂoor, the
   string detaches from the axle.
                                                  Flywheel

                             Axle

                                                                 d

                                    m
                                                                        Rigid support
             Rigid support

                                             h
                                                         E
                                        CS
                 HB
  From the instant when the mass touches the ﬂoor (taken as t = 0), the ﬂywheel continues to
  rotate, adding another N number of rotations before coming to rest in time t = T . The ﬁgure is
  not to scale.
   (a) [5 marks] Derive an expression for the moment of inertia I of the ﬂywheel in terms of
       m, N, T , and other known parameters.

         Solution: Using conservation of energy
                                                  1      1
                                        mgh = nW + mv 2 + Iω 2                               (1.1)
                                                  2      2
         Also,
                                                       1
                                               N W = Iω 2                                    (1.2)
                                                       2
                                                        Iω 2
                                                   W =                                       (1.3)
                                                         2N
         Substituting Eq.(1.3) in Eq.(1.1), we get
                                                1        1     1
                                    mgh = n.      Iω 2 + mv 2 + Iω 2                         (1.4)
                                              2N         2     2
         Simplifying above equation we get,
                                               2mgh         md2
                                        I=          n
                                                       −       n
                                                                                (1.5)
                                             ω2 1 + N     4 1+ N
         where ω = 4πN
                     T
         Final answers written in the form of W are also accepted.

Page 3       INPhO 2025                    Questions

   (b) [3 marks] The following data has been obtained in this experiment.

                                      m (gm)    N         T (sec)

                                      150.0     145.25    190.0

                                      200.0     200.00    225.5

                                      250.0     238.50    235.5

         Calculate the value of I for n = 16, and h = 139 cm.

          Solution: From the given data, we can calculate the respective value of the moment of
          inertia I1 , I2 , I3 are 3.99 × 105 gm-cm2 , 4.06 × 105 gm-cm2 , 3.94 × 105 gm-cm2 . The mean
          value of the moment of inertia is 4.00 × 105 gm-cm2 .

                                                     E
                                  CS
            HB

Page 4       INPhO 2025                    Questions          Last four digits of Roll No.:

INPhO 2025 Problem 1


[User]
A photon of energy E bounces between two mirrors separated by a distance L.

      (a) If the mirrors are slowly moved together to a distance L/2, find the final energy E by consid-
          ering the blueshift the photon experiences at every collision.

     (b) Check this result agrees with the adiabatic theorem.

        (c) Using your result, infer the value of γ for a one-dimensional photon gas. Can you also infer
            the value of γ for a d-dimensional photon gas?

Solution:
(a) If the left mirror has speed v, then the Doppler shift factor on every reflection is
         1 + 2v/c, so ∆E = (2v/c)E because E = ℏω. We also have ∆x = −(2x/c)v, which implies

                                                     ∆E   E
                                                        =− .
                                                     ∆x   x
           Treating this as a differential equation dE/dx = −E/x shows that Ex is conserved. (Note that
           the adiabatic theorem works fine in this case, because the photon is always moving quickly
           relative to the walls, so every collision has a small effect.)
           Another valid, equivalent approach would be to note that the energy gets an overall Doppler
           shift of (1 + 2v/c)N where N is the number of collisions with the moving wall, and compute
           N directly.

     (b) The adiabatic invariant is           I
                                                            E
                                                   p dx =     (2L) ∝ Ex
                                                            c
           giving the same result, using E = pc.

        (c) The pressure exerted by each photon is

                                                     ∆p   2E/c  E
                                              P =       =      = .
                                                     ∆t   2L/c  x

           Therefore, P x2 is constant which gives γ = 2. This is exactly as expected, since Cv = kT .
           In d dimensions, much of the same reasoning holds. The quantity Ex is still invariant,
           where x is the side length of, say, a cube. However, the pressure is now P ∝ E/V . Then
           P V x = P V 1+1/d is constant, so γ = 1 + 1/d. For example, when d = 3 we have γ = 4/3, as
           we found in T1 and T2.
           Note that the reasoning here works equally well if you don’t know about photons, because
           electromagnetic waves redshift and blueshift in the same way. By using this reasoning, i.e. by
           thinking about how successive Doppler shifts change the frequency of light, Wien was able to
           derive the original form of Wien’s displacement law.

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


[User]
Here’s another quick application of the conserved Jz identified in problem 15. As
    discussed in E8, electron orbits can be modified in a magnetic field, leading to diamagnetism. Many
    textbooks try to motivate this by considering forces on classical electrons, but such arguments don’t
    actually work: it can be shown that for classical systems in thermal equilibrium, diamagnetic effects
    always cancel out. A legitimate derivation requires some quantum mechanics.
       In X1, we discussed how the electron orbits in a hydrogen atom have Lz = nℏ. Let’s suppose
    the states n = ±1 are occupied, corresponding to electrons of charge q and mass m performing a
    circular orbit of radius r in opposite directions. The magnetic moments associated with these orbits
    cancel. But in the presence of a uniform magnetic field B, perpendicular to the plane of the orbit,
    the quantization condition becomes Jz = nℏ, and the magnetic moments no longer cancel.
       Find an approximate expression for the net magnetic moment, in terms of q, m, r, and B.
    Assume B is small, so that the magnetic force is small compared to the electrostatic force.

Solution:
This problem was inspired by this paper. Following problem 15, we note that for the
    n = 1 orbit,

                                           Jz = Lz + qBr′2 = ℏ

            ′
    where r is the orbit radius when the magnetic field is on. In the absence of a magnetic field, we
    have Lz = ℏ, so the change in Lz is

                                              ∆Lz = − qBr′2 .

                                       ′
    Since the magnetic field is weak, r ≈ r, and since ∆Lz is proportional to B, which is already small,
    we can neglect the difference between r and r′ . Finally, using a result from E5, we have

                                                 q         q 2 Br2
                                        ∆µz =      ∆Lz = −         .
                                                2m           4m
    For the n = −1 orbit, the change in magnetic moment has the same sign, giving a total of

                                                       q 2 Bz r2
                                              µz = −             .
                                                          2m


[User] Juku found a hemispherical rock sticking up from the ground. Measuring its circumference with a tape he got a result of $a=\SI{2,4}{m}$. Next he took a match box and started to slowly move it down from top of the hemisphere, until the box finally slid off the rock. Measuring along the hemisphere with a tape, Juku found that the distance from the tip of the rock to the point where the box slid off was $b=\SI{20}{cm}$. What was the coefficient of friction between the rock and the match box?

Solution:
Olgu libisema hakkamise hetkel pinna kaldenurk $\alpha$. Mõtleme esmalt, kuidas on seotud see $\alpha$ väärtus hõõrdeteguriga $\mu$. Topsile mõjuva raskusjõu $mg$ jagame pinnaga risti olevaks komponendiks $F_{\bot}=mg\cos\alpha$ ning pinnaga paralleelseks komponendiks $F_{||}=mg\sin\alpha$. Kivi pind avaldab topsile toereaktsiooni $N=F_\bot$ ning maksimaalset hõõrdejõudu $F_h=\mu N$. Tops libiseb maha, kui $F_{||}>F_h$. Kriitilisel hetkel saame võrrandi $F_{||}=F_h$ ehk
\[ \mu \cos \alpha = \sin\alpha, \quad \text{millest} \quad \tan\alpha = \mu.\]

Kaldenurk $\alpha$ on ühtlasi võrdne vertikaali ja kivi keskpunktist libisemispaika tõmmatud joone vahelise nurgaga ja suhtub täisringi $360^\circ$ samuti kui kaarepikkus $b$ suhtub ümbermõõtu $a$, $\alpha = 360^\circ \!\cdot\! b/a$. Seega $\mu = \tan (360^\circ \! \cdot\! b/a)$ ning arvuliseks vastuseks saame
\[\mu = \tan (30^\circ) = \frac{1}{\sqrt{3}} \approx \SI{0,58}{}.\]

Kivi

Juku avastas maast välja turritamas poolkerakujulise kivi. Mõõtes mõõdulindiga
selle ümbermõõdu, sai ta tulemuseks $a=\SI{2,4}{m}$. Edasi võttis ta taskust tikutopsi ja
hakkas seda poolkera tipust alates natukesehaaval allapoole liigutama, kuni
lõpuks tikutops kivilt maha libises. Mõõdulindiga mööda poolkera mõõtes sai ta
tipu ja libisemispaiga vaheliseks kauguseks $b=\SI{20}{cm}$. Kui suur oli kivi
ja tikutopsi vaheline hõõrdetegur?

Tikk hakkab poolkera pealt maha libisema siis, kui raskusjõu pinnaga paralleelne komponent ületab hõõrdejõu.


[User]
8. A conducting wire frame of single turn in the shape of a rectangle ABCD (sides AB = a,            [7]
    BC = b) is free to rotate about the side AB which is horizontal. Initially, the frame is held
    in a horizontal plane and a steady current i (clockwise as seen from above) is switched on in
    the wire and then the frame is released (still free to rotate about AB). Find the magnitude
    and direction of the minimum uniform magnetic field necessary to keep the frame horizontal.
    The origin of the coordinate system is at A and AB lies along the direction of the +x axis.
    Take the mass of the wire per unit length as λ and the acceleration due to gravitational
    field to be g k̂. Here k̂ is unit vector in +z direction.

           **** END OF THE QUESTION PAPER ****

                 HOMI BHABHA CENTRE FOR SCIENCE EDUCATION
                        Tata Institute of Fundamental Research
                    V. N. Purav Marg, Mankhurd, Mumbai, 400 088

Solution:
8. A conducting wire frame of single turn in the shape of a rectangle ABCD (sides AB = a,            [7]

            HB
    BC = b) is free to rotate about the side AB which is horizontal. Initially, the frame is held
    in a horizontal plane and a steady current i (clockwise as seen from above) is switched on in
    the wire and then the frame is released (still free to rotate about AB). Find the magnitude
    and direction of the minimum uniform magnetic field necessary to keep the frame horizontal.
    The origin of the coordinate system is at A and AB lies along the direction of the +x axis.
    Take the mass of the wire per unit length as λ and the acceleration due to gravitational
    field to be g k̂. Here k̂ is unit vector in +z direction.

                                              λ(a + b)g
     Solution: Minimum magnetic field =                 which is in the direction of +y or −y
                                                 ia
     axis.

           **** END OF THE QUESTION PAPER ****

                 HOMI BHABHA CENTRE FOR SCIENCE EDUCATION
                        Tata Institute of Fundamental Research
                    V. N. Purav Marg, Mankhurd, Mumbai, 400 088

INPhO 2014 Problem 8


[User]
Consider two conducting spheres of radius r separated by a distance a ≫ r, with
    total charges ±Q. The spheres can be thought of as the two plates of a capacitor.
      (a) Find a simple approximation for the capacitance C, valid when a ≫ r.

In reality, the exact capacitance of this system can be written as an infinite series in r/a. Let’s
consider two ways of finding the corrections to the capacitance.

 (b) By considering the energy U = Q2 /2C of the system, find the first nontrivial correction to C.
    (c) Alternatively, we can think about the charge distributions on the spheres. If we start with
        a “zeroth-order” uniform charge density on each sphere, it will induce a “first-order” image
        charge in the other sphere, which will in turn induce “second-order” image charges, and so
        on. We can then compute C = Q/∆V by summing up all the image charges, and the total
        voltage difference they produce. Using this approach, find the first nontrivial correction to C.
 (d) ⋆ It turns out that the quantity 1/C is a bit nicer than C. At what order in r/a does the
     second nontrivial correction to 1/C appear?

     Example 6

     Find the charge distribution on a thin conducting disc of radius R and total charge Q.

     Solution
     In general, there are very few situations where the charge distribution on a conductor can
     be found explicitly. As you’ve seen, some of the simplest examples can be solved with image
     charges. Some more complex, two-dimensional examples can be solved with a mathematical
     technique called conformal mapping. And this special example can be solved with a neat trick.

     Consider a uniformly charged spherical shell centered on the origin, and consider a point P
     inside the shell, on the xy plane. The electric field at point P is zero, by the shell theorem.
     Recall that in the usual proof of the shell theorem, one draws two cones opening out of
     P in opposite directions. The charges contained in each cone produce canceling electric fields.

     Now imagine shrinking the spherical shell towards the xy plane, so it becomes elliptical. The
     crucial insight is that the shell theorem argument above still works, for points on the xy
     plane. When we squash the shell all the way down to the xy plane, it becomes a disc, with
     zero electric field on it. This is thus a valid charge distribution for a disc-shaped conductor,
     and by the uniqueness theorem, it’s the only one.

     By√keeping track of how much charge gets squashed to radius [r, r + dr], we find σ(r) ∝
     R/ R2 − r2 , and fixing the proportionality constant gives

                                                      Q
                                         σ(r) =      √        .
                                                  4πR R2 − r2
     You can also show this by taking the c, ϵ → 0 limit of the “third shell theorem” in M6. Note
     that this is the surface charge density on each side of the thin disc, so if you wanted the limit
     of an infinitely thin disc, you should double the answer.

4      Electrical Conduction
We now leave the world of electrostatics and consider magnetostatics, the study of steady currents.

       Idea 6
       In a conductor with conductivity σ, the current density is

                                                   J = σE.

       Alternatively, E = ρJ where ρ is the resistivity. The current and charge density satisfy
                                                           ∂ρ
                                                ∇·J=−         .
                                                           ∂t
       The current passing through a surface S at a given time is
                                                 Z
                                             I=     J · dS.
                                                      S

       Since J ∝ E, we have Ohm’s law V = IR, where V is the voltage drop across the resistor.
       The power dissipated in a resistor is P = IV . The resistance R adds in series, while 1/R
       adds in parallel.

[2] Problem 21 (HRK). A battery causes a current to run through a loop of wire.

      (a) Suppose the wire makes a sharp corner. How do the charges know to turn around there?

      (b) A copper wire with conductivity σ is joined to an iron wire with conductivity σ ′ < σ. For
          the current in both sections to be the same, the electric field in the iron wire must be higher.
          How does that happen?

    For more about surface charges in circuits, see this paper and this paper. For a great visualization
    of how the charge and current configuration in a real circuit is set up over time, see this video.

Solution:
(a) The potential of a single sphere, relative to infinity, is V = Q/4πϵ0 r. Hence the
         two spheres, at this level of approximation, have potentials ±Q/4πϵ0 r, and the capacitance is
                                                       Q
                                               C=           = 2πϵ0 r.
                                                    V1 − V2
          This is ignoring any interaction between the charges on different spheres.

     (b) Each sphere has an energy Q2 /8πϵ0 r due to its own field, so the simplest correction is to
         account for the electrostatic interaction between them, treating them as approximately point
         charges. We therefore have
                                                Q2     Q2     Q2
                                        U=           +      −       + ...
                                               8πϵ0 r 8πϵ0 r 4πϵ0 a
          which gives
                                         C = 2πϵ0 r (1 + r/a + O(r2 /a2 )).

 (c) Let’s review the strategy. If we put a uniform charge on one sphere, then by the shell theorem,
     its effect on the second sphere is the same as a point charge at the first sphere’s center. This
     induces an image charge on the second sphere. The same logic applies in reverse, with uniform
     charge on the second sphere inducing an image charge on the first sphere. The image charges
     are smaller than the original ones by a factor of r/a, by the result of problem 2.
     However, this isn’t a solution to the problem, because these “first-order” image charges on
     each sphere in turn induce “second-order” image charges in the other sphere, which are smaller
     by another factor of r/a, and so on. Once we sum up an infinite series of image charges, we
     get the true charge configuration, from which we can compute the exact capacitance, as a
     series in r/a. The net charge on each sphere is the sum of all its image charges.
     Specifically, let the total charge and voltage on the spheres be ±Q and ±V . Both of these
     quantities can be expanded in a power series in r/a,
                          Q = q0 + q1 + q2 + . . . ,    V = V0 + V1 + V2 + . . .
     and the capacitance is simply C = Q/(V − (−V )) = Q/2V .
     To start, we put a “zeroth-order” charge of q0 at the center of the positive sphere. This charge
     induces a voltage V0 = q0 /(4πϵ0 r) on that sphere. Now, the negative sphere’s −q0 induces
     an image charge q1 = q0 r/a in the positive sphere. By construction, the −q0 on the negative
     sphere and the +q1 image charge on the positive sphere together yield zero potential on the
     positive sphere, so V1 = 0. (This is a lucky result, which won’t continue at higher orders.) So
     the capacitance with the first correction is simply
                         Q     q0 + q1 + O(r2 /a2 )
                   C=      =                          = 2πϵ0 r (1 + r/a + O(r2 /a2 )).
                        2V   2 (V0 + V1 + O(r2 /a2 ))
     in agreement with part (b).

 (d) As we’ve seen, the answer is a power series in r/a, so we would guess that the next term is of
     order (r/a)2 . However, the next term is actually of order (r/a)4 .
     This is difficult to see with the image charge method of part (c). If you work it out, you’ll
     find that the second order and third order corrections to 1/C cancel out, in a complicated
     way. On the other hand, it’s intuitive if you consider the energy, as in part (b). This is the
     more natural quantity, since U ∝ 1/C.
     We know that the leading correction to the charge distribution is the image charge q1 = q0 r/a,
     which is off center by r2 /a. Thus, each sphere can be regarded as a point charge Q at its
     center, plus a dipole moment p1 ∼ Qr3 /a2 , plus higher order corrections. This dipole interacts
     with the point charge field of the other sphere with a potential energy
                                                       Qr3 Q   Q2 r3
                                     ∆U ∼ p1 · E ∼           ∼
                                                       a2 a2    a4
     which is order (r/a)4 smaller than the leading term in U , as promised.
     For much more about this problem, see this paper. The result we derived in part (b) matches
     its equation (2.2), upon identifying Qa → Q, Qb → −Q, a, b → r, and c → a, and using cgs
     units where 4πϵ0 = 1. The third correction shows up at order (r/a)6 , and corresponds to a
     dipole-dipole interaction. As noted in the paper, Maxwell was so interested in this system
     that he calculated 1/C out to 22nd order!

    Example 6

    Find the charge distribution on a thin conducting disc of radius R and total charge Q.

    Solution
    In general, there are very few situations where the charge distribution on a conductor can
    be found explicitly. As you’ve seen, some of the simplest examples can be solved with image
    charges. Some more complex, two-dimensional examples can be solved with a mathematical
    technique called conformal mapping. And this special example can be solved with a neat trick.

    Consider a uniformly charged spherical shell centered on the origin, and consider a point P
    inside the shell, on the xy plane. The electric field at point P is zero, by the shell theorem.
    Recall that in the usual proof of the shell theorem, one draws two cones opening out of
    P in opposite directions. The charges contained in each cone produce canceling electric fields.

    Now imagine shrinking the spherical shell towards the xy plane, so it becomes elliptical. The
    crucial insight is that the shell theorem argument above still works, for points on the xy
    plane. When we squash the shell all the way down to the xy plane, it becomes a disc, with
    zero electric field on it. This is thus a valid charge distribution for a disc-shaped conductor,
    and by the uniqueness theorem, it’s the only one.

    By√keeping track of how much charge gets squashed to radius [r, r + dr], we find σ(r) ∝
    R/ R2 − r2 , and fixing the proportionality constant gives

                                                     Q
                                        σ(r) =      √        .
                                                 4πR R2 − r2
    You can also show this by taking the c, ϵ → 0 limit of the “third shell theorem” in M6. Note
    that this is the surface charge density on each side of the thin disc, so if you wanted the limit
    of an infinitely thin disc, you should double the answer.

4    Electrical Conduction
We now leave the world of electrostatics and consider magnetostatics, the study of steady currents.

    Idea 6
    In a conductor with conductivity σ, the current density is
                                                 J = σE.
    Alternatively, E = ρJ where ρ is the resistivity. The current and charge density satisfy
                                                     ∂ρ
                                             ∇·J=−      .
                                                     ∂t
    The current passing through a surface S at a given time is
                                              Z
                                          I=     J · dS.
                                                   S

       Since J ∝ E, we have Ohm’s law V = IR, where V is the voltage drop across the resistor.
       The power dissipated in a resistor is P = IV . The resistance R adds in series, while 1/R
       adds in parallel.

[2] Problem 21 (HRK). A battery causes a current to run through a loop of wire.

      (a) Suppose the wire makes a sharp corner. How do the charges know to turn around there?

      (b) A copper wire with conductivity σ is joined to an iron wire with conductivity σ ′ < σ. For
          the current in both sections to be the same, the electric field in the iron wire must be higher.
          How does that happen?

    For more about surface charges in circuits, see this paper and this paper. For a great visualization
    of how the charge and current configuration in a real circuit is set up over time, see this video.

    Solution. (a) The first charges to make it there don’t; they just stop at the surface of the wire,
         due to the attraction from the positively charged nuclei. Once this charge builds up at the
         kink, it repels the next electrons so that they automatically turn around. This typically occurs
         extremely quickly, as the relevant timescale is the RC of the wire and C is tiny. The amount
         of charge required is very small, less than a few hundred electrons.

      (b) It’s the same story as part (a). The first charges to reach the iron will start moving slower,
          because the fields are the same. This then causes a buildup of charge at the interface between
          them, which increases the field in the iron and decreases the field in the copper. In the steady
          state, the current densities in both are equal. Again, this occurs very quickly and requires
          very little charge.


[User] An alpinist with a mass of $m=\SI{75}{kg}$ is fastened to an elastic rope with a length of $L=\SI{6}{m}$. The other side of the rope is attached to a cliff. After climbing to a height of 6 meters from the attachment point the alpinist falls. Find the rope’s maximal elasticity coefficient $k$, knowing that a maximal pulling force of the rope that a human can tolerate is $T=25mg$. Do not account for air resistance.

Solution:
Alpinist kukub $2L+x$ võrra, kus $x$ on nööri pikenemine. Jõudes kukkumise kõige madalamasse punkti, kus kineetiline energia puudub, on gravitatsiooniline potentsiaalne energia $mg(2L+x)$ üle läinud vedru potentsiaalseks energiaks $kx^2/2$. Seega energia jäävuse seadus väljendub kujul
\[mg(2L+x)=\frac{kx^2}{2}.\]
Maksimaalse nööri pinge seosest $T=kx=25mg$ avaldame $x=25mg/k$ ning asendades selle eelnevasse võrrandisse, saame
\[4mgL+\frac{50m^2g^2}{k}=\frac{25^2m^2g^2}{k}.\]
Lihtsustamiseks korrutame võrrandit $k$-ga ning jagame $4mgL$-ga, saame
\[k=\frac{(25^2-50)mg}{4L}\approx \SI{18}{kN/m}.\]
({\em Huvitav on ka pikenemine $x$ välja arvutada, saame} $x\approx \SI{1}{m}$.)

Alpinist

Alpinist massiga $m=\SI{75}{kg}$ on kinnitatud elastse nööri külge pikkusega
$L=\SI{6}{m}$. Nööri teine ots on kinnitatud kalju külge. Olles roninud 6 meetri
kõrgusele kinnituskohast, ta kukub. Leidke, kui suur võib olla ülimalt nööri
elastsustegur $k$, teades, et suurim nööri tõmbejõud, mida inimene talub, on
$T=25mg$. Õhutakistust ärge arvestage.

Jõudes kukkumise madalaimasse punkti, on alpinisti potentsiaalne energia läinud üle vedru potentsiaalseks energiaks. Pannes vastava seose kirja, on võimalik leida nööri pikenemise $x$.


[User]
A rotor consists of two particles of mass m connected by a rigid rod of length L.

     (a) Find the energy levels if the particles are not identical.

     (b) Find the energy levels if the particles are identical. (Hint: recall that the closed loop integrals
         in the previous ideas are over paths that take the system back to its original state.)

    You might find it disturbing that the result is so different if the particles are or aren’t completely
    identical, but it’s a well-verified fact about molecular rotational energy levels. For example, if the
    two atoms in the molecule had nuclei of different isotopes, we measure the result of part (a), and if

                   01h
    they were the same isotope, we measure the result of part (b).

Solution:
(a) We see that the angular momentum Lz is constant, so the quantization condition
                                                   1 2      2 2
         says Lz · 2π = nh, or Lz = nℏ. Then, E = 2I Lz = nmLℏ2 .
     (b) The point here is that after just a π rotation, the system is back in its original state, because
                                                                                2 ℏ2
         the particles are identical. Thus, Lz · π = nh, or Lz = 2nℏ, or E = 4n

                    01h
                                                                                mL2
                                                                                .


[User]
4. Magnetic field (6 points)

MagneticfieldwithinductanceB (paralleltothez-axis)fillsthe
region x2
+ y2
< R2
. Let us consider an electron of velocity
v = RBe/m (wheree isitschargeandm itsmass).
1) Sketchthetrajectoryoftheparticle,ifinitially,itmovesalong
theliney = 0 towardstheregionfilledwithmagneticfield.
2) How long time does the electron spend inside the magnetic
field?
3) Now,letusconsiderthesituation,wheninitiallytheelectron
movesalongtheliney = a (a < R). Findtheangleα,bywhich
theelectronisinclinedafterpassingthroughmagneticfield.

Solution:
4. Magnetic field

1) Sincetheradiusofthecyclotronorbitisequaltotheradius
oftheregionR, thetrajectoryisgivenbythecurveDABE in
theFigure(AB isacirclefragment).
2) Circularpartofthetrajectoryisaquarterofthefullcircle,
sot = πR/2v.
3) LetO′
bethecentreofthecircularorbitoftheelectronandB
theintersectionpointofthetrajectorywiththeregionbound-
ary. The polygon COBO′
is rhomb, because all the sides are
equaltoR. So,thelineBO isvertical(becauseO′
C isvertical
and BO is parallel to it). Hence, the inclination angle of the
electronis
α = ∠CO′
B = ∠AOB + ∠AOC =
π
2
− arcsin
a
R
.

Magnetic field


[User]
After reading the chapter, do the following as well as you can without references.
    (You can peek if you need to, but try to do that as little as possible.)
      (a) Write down the fundamental particles of the Standard Model, along with their electric charges.

      (b) Order the particles from lightest to heaviest.

        (c) Which particles make up most of what you see in the everyday world?

      (d) Which particles participate in the strong interaction?

        (e) Which particles participate in the weak interaction?
[3] Problem 16 (Griffiths 1.19). Your roommate is a chemistry major. She knows all about protons,
    neutrons, and electrons, and she sees them in action every day in the laboratory. But she is skeptical
    when you tell her about positrons, muons, neutrinos, pions, quarks, and intermediate vector bosons.
    Explain to her why each of these play no direct role in chemistry.
    Olympiad questions about particle colliders boil down to questions from E4, E7, and R2, so they

                           01^
    should be fairly straightforward if you know the principles.

Solution:
(a) Quarks: up, charm, top have charges of +2/3, and down, strange, bottom have
         charges of −1/3.
           Leptons: electron, muon, tau have charges of −1, and their neutrinos have no charge.
           The gluon, photon, Z boson, and Higgs boson have no charge, and the W± boson has a charge
           of ±1.

      (b) Photon/gluon (massless), electron neutrino, muon neutrino, tau neutrino, electron, up quark,
          down quark, strange quark, muon, charm quark, tau, bottom quark, W boson, Z boson, Higgs
          boson, top quark.

        (c) The up and down quarks in nucleons, electrons, and photons.

      (d) Gluons mediate the strong interaction, which affects quarks and gluons.

        (e) W/Z bosons mediate the weak interaction, which affects quarks and leptons (the electron,
            muon, tau, and corresponding neutrinos).

[3] Problem 16 (Griffiths 1.19). Your roommate is a chemistry major. She knows all about protons,
    neutrons, and electrons, and she sees them in action every day in the laboratory. But she is skeptical
    when you tell her about positrons, muons, neutrinos, pions, quarks, and intermediate vector bosons.
    Explain to her why each of these play no direct role in chemistry.

    Solution. Positrons will annihilate after contacting electrons, muons are unstable and decay in a
    few microseconds, neutrinos are too small and have no charge so they’re too elusive to be detected,
    pions are even more unstable than muons, quarks are locked inside protons and neutrons (they
    won’t exist on their own since they’ll be unstable that way), and the bosons have a half life of
    around 3 × 10−25 s.

    Olympiad questions about particle colliders boil down to questions from E4, E7, and R2, so they

                     01^
    should be fairly straightforward if you know the principles.


[User]
Q2-1
Theory

                                                                                Oﬃcial (English)

A Mechanical Model for Phase Transitions1
A ring of radius 𝑅 has a bead of negligible size and mass 𝑚 threaded on it. The ring is set rotating
about its vertical diameter with angular velocity 𝜔 as shown in Fig. 1. Alongside this, there is an
opposing force on the bead, 𝐹𝑓 , which is proportional to the normal reaction 𝑁 , and is given by
𝐹𝑓 = 𝑓𝑘𝑁 where 𝑓 is 1 or −1. You are advised to employ polar coordinates {𝑟, 𝜃}. As far as possible
express your answers in terms of 𝜔𝑐 = √𝑔/𝑅 where 𝑔 is the magnitude of the acceleration due to gravity.

In a phase transition the free energy, 𝑉 (ℳ) [mechanical equivalent is potential energy] depends on the
magnetization ℳ as follows
                                        𝑉 (ℳ) = 𝑎(𝑇 )ℳ2 + 𝑏(𝑇 )ℳ4

where 𝑇 → temperature, 𝑏(𝑇 ) > 0 and 𝑎(𝑇 ) changes sign with temperature. We attempt to understand
the phenomenon of phase transition using the above mentioned model.
Note 1: For circular motion of radius 𝑅, the velocity in polar coordinates is 𝑟 ⃗̇ = 𝑅𝜃𝜃̇ ̂ and the acceleration
is 𝑟 ̈⃗ = −𝑅𝜃 ̇2 𝑟 ̂ + 𝑅𝜃 ̈𝜃.̂ Here 𝑟 ̂ and 𝜃 ̂ are unit vectors in the radial and the tangential directions respectively.
Note 2: The direction of the opposing force say 𝐹𝑓⃗ will be denoted by 𝑓. Here 𝑓 = +1 if the bead is moving in
the counter-clockwise direction (of increasing) 𝜃 and 𝑓 = -1 if the bead is moving in the clockwise direction
𝜃, e.g. 𝑓 = 𝑠𝑔𝑛(𝜃)̇ where 𝑠𝑔𝑛 is +1 or -1 depending on whether its argument is positive or negative.
Note 3: You may ﬁnd the expansions

                             sin(𝜃)     =   𝜃 − 𝜃3 /6 + ..
                            cos(𝜃) =        1 − 𝜃2 /2 + 𝜃4 /24 + ..
                                    𝑛
                          (1 + 𝑥)       =   1 + 𝑛𝑥 + 𝑛(𝑛 − 1)𝑥2 /2 + 𝑛(𝑛 − 1)(𝑛 − 2)𝑥3 /6 + ..

(where 𝜃 is in radians and |𝑥| << 1 ) useful for some parts of the problem.
  1 Sitikantha Das (IIT Kharagpur) and Pramendra Ranjan Singh (Principal, Narayan College, J.P. University) were the principal

authors of this problem. The contributions of the Academic Committee, Academic Development Group, and the International
Board are gratefully acknowledged.

                                                                           Q2-2
Theory

                                                                                Oﬃcial (English)

                                 Fig. 1: The bead on a rotating ring.

In what follows we shall understand the dynamics of the bead in the frame of the rotating ring and for
angles in the range −𝜋/2 < 𝜃 < 𝜋/2. The free body diagram of the bead is shown in Fig.2. Neglect all
forces other than the ones shown in the free body diagram.

                                    Fig. 2: The free body diagram.

    A.1     Write down the equations of motion for the radial 𝐹𝑟 and the tangential 𝐹𝜃 com-   0.5pt
            ponents of the force on the bead. Assume 𝜃 increasing in the counter-clockwise
            direction.

For the following parts B.1 to B.9 assume 𝑘 = 0.

                                                                              Q2-3
Theory

                                                                                Oﬃcial (English)

    B.1    State the relation between the equilibrium angle(s) 𝜃0 in terms of { 𝜔, 𝜔𝑐 }.          1.0pt

    B.2    Qualitatively sketch 𝜃0 (𝑦-axis) as a function of 𝜔/𝜔𝑐 (𝑥-axis).                       0.5pt

    B.3    Qualitatively sketch the magnitude of the normal reaction force on the bead as         0.5pt
           a function of 𝜔/𝜔𝑐 at stable equilibrium.

    B.4    We deﬁne the potential energy corresponding to the tangential force 𝐹𝜃 ,               1.0pt
           namely
                                                   1 𝑑
                                           𝐹𝜃 = −       𝑉 (𝜃)
                                                   𝑅 𝑑𝜃
           with the zero of potential energy at 𝜃 = 0. If 𝑉 (𝜃) is expressed as 𝑃 + 𝑄 cos(𝜃) +
                2
           𝑆 sin (𝜃), obtain 𝑃 , 𝑄, 𝑆.

    B.5    One can expand 𝑉 (𝜃) for small 𝜃 and express it as 𝑉 (𝜃) = 𝑎(𝜔) 𝜃2 +𝑏(𝜔) 𝜃4 . Obtain   1.0pt
           the coeﬃcients 𝑎(𝜔) and 𝑏(𝜔).

    B.6    Make representative plots of 𝑉 (𝜃) versus 𝜃 for values of 𝜔/𝜔𝑐 just less than 1.0      1.0pt
           (e.g. say 0.9) and 𝜔/𝜔𝑐 large, say 5.0. Note that only qualitative sketches and no
           detailed calculations for the plots are required.

    B.7    Landau theory of second order phase transitions can be used to demarcate               1.0pt
           simple magnetic systems into two phases. For temperatures 𝑇 greater than
           the critical temperature 𝑇𝑐 the system is paramagnetic. For 𝑇 < 𝑇𝑐 the system
           is ferromagnetic and the magnetization ℳ is given by

                                   ℳ(𝑇 ) = ℳ0 (1 − 𝑇 /𝑇𝑐 )1/2      𝑇 < 𝑇𝑐

           Let us denote the exponent 1/2 by 𝛽. Compare this behaviour with the bead
           problem discussed above. What are the analougues of ℳ, 𝑇𝑐 , 𝑇 /𝑇𝑐 in our case?
           What is the equivalent value of 𝛽 in our case?

    B.8    Determine the angular frequency of oscillation Ω0 of the bead when it is dis-          1.0pt
           turbed from its “equilibrium” position 𝜃0 . Note that for small oscillations

                                                   1 √ 𝑉 ″ (𝜃0 )
                                            Ω0 =
                                                   𝑅      𝑚

    B.9    Qualitatively sketch Ω0 as a function of 𝜔.                                            1.0pt

For the following parts C.1 to C.2 𝑘 ≠ 0.

                                                                           Q2-4
Theory

                                                                                Oﬃcial (English)

  C.1    Take 𝑓 = 1 and express 𝑘 = tan 𝛼. We may express the condition for the equilib-      1.0pt
         rium angle(s) 𝜃0 as
                                              2
                                           𝜔     tan(𝑥)
                                        ( ) =
                                          𝜔𝑐     sin(𝑦)
         Obtain 𝑥 and 𝑦.

  C.2    It is given that 𝑓 =1 and 𝑘 = 0.05. Obtain the equilibrium angles 𝜃0 , if any, for   0.5pt
         the following cases:
             1. 𝜔/𝜔𝑐 = 0.50
             2. 𝜔/𝜔𝑐 = 0.70

Solution:
Solutions

                                                                                A2-1       Oﬃcial (English)

A Mechanical Model for Phase Transitions1
General Grading Guidelines
When student’s solutions are correct and s/he also show how solutions were obtained, the stduent gets
full credit. The scheme oulined below is helpful if the student’s answers are partially correct. Atten-
tion will be paid to the detailed solution so, if the ﬁnal answer is correct but it is obtained by incorrect
method(s) then no credit will be given. Alternative solutions may exist and will be given due credit.
Partial or full outcomes obtained for later sections in the problem which are incorrect solely because of
errors being carried forward from previous sections, but are otherwise reasonable, will not be further
penalized. For example a dimensioanlly wrong answer when carried forward will not get any credit in the
subsequent sections. A numerically wrong evaluation when carried forward will get credit in subsequent
sections unless the numerical answer is patently wrong (e.g. the value of g is 981 m/sec2 ! )
Incorrect or no labeling of an axis is penalized by -0.1 points
The numerical answer (i) must be correct to +/- 10% AND (ii) must respect signiﬁcant ﬁgures.
It maybe noted that NO micro-marking scheme takes care of all contingencies. A certain amount
of discretion rests with and a certain level of judgement is invested in the academic committee.

  A.1 (0.5 pt)
  Equations of motion

  The radial component 𝐹𝑟 yields:
                                                                                2
                                      𝑚𝑅𝜃 ̇2   = 𝑁 − 𝑚𝑔 cos(𝜃) − 𝑚𝑅 sin (𝜃)𝜔2                                           (1)

                                                                                [0.2]
  The tangential component 𝐹𝜃

                             𝑚𝑅𝜃 ̈ = 𝑚𝑅 sin(𝜃) cos(𝜃)𝜔2 − 𝑚𝑔 sin(𝜃) − sgn (𝜃)̇ 𝑘𝑁                                       (2)
                              OR
                             𝑚𝑅𝜃 ̈ = 𝑚𝑅 sin(𝜃) cos(𝜃)𝜔2 − 𝑚𝑔 sin(𝜃) − 𝑓𝑘𝑁 (𝑓 = 1)                                       (3)

                                                                                [0.3]

  No points if equations not written using radial and tangential components.

  1 Sitikantha Das (IIT Kharagpur) and Pramendra Ranjan Singh (Principal, Narayan College, J.P. University) were the principal

authors of this problem. The contributions of the Academic Committee, Academic Development Group, and the International
Board are gratefully acknowledged.

Solutions

                                                                                A2-2      Oﬃcial (English)

 B.1 (1.0 pt)
 Equilibrium angle(s)

 We set 𝑘 = 0 in the equation for the tangential component of the force. Thus

                                𝑚𝑅𝜃 ̈ =    𝑚𝑅 sin(𝜃) cos(𝜃)𝜔2 − 𝑚𝑔 sin(𝜃)                              (4)

 For equilibrium we set 𝜃0̈ = 0 in the above equation. Then 𝜃0 = 0 is an equilibrium angle for all values
 of 𝜔
                                                                                [0.4]
 The other values are given by
                                                     𝑔     𝜔2
                                           cos 𝜃0 = 2 = 𝑐2                                             (5)
                                                    𝜔 𝑅    𝜔
                                                                                [0.3]

                                                            𝜔𝑐2
                                            𝜃0 = ± ∣cos−1       ∣                                      (6)
                                                            𝜔2
                                                                                [0.1]
 with values of 𝜃0 between -𝜋/2 to 𝜋/2. The ± indicates that there are two equivalent positions. () The
 bead could rise on either side of the axis shown in the ﬁgure depicted in the problem. Note that for
 𝜔 < 𝜔𝑐 , Eq. (5) implies cos 𝜃0 > 1. This is clearly unphysical. A little reﬂection will convince us that
 𝜃0 = 0 for 𝜔 < 𝜔𝑐 .
                                                                                [0.2]

 B.2 (0.5 pt)
 Sketch of 𝜃0 .

 𝜃0     0

                  1
                              𝜔/𝜔𝑐
                                                                                [shape correct: 0.2]
                                                                            [only if both branches: 0.3]

Solutions

                                                                              A2-3      Oﬃcial (English)

 B.3 (0.5 pt)
 Sketch of the magnitude of the normal reaction

 𝑁

      𝑚𝑔

                             1
                             𝜔/𝜔𝑐
                                                                                [𝜔 < 𝜔𝑐 ∶ 0.2]
                                                                                [𝜔 > 𝜔𝑐 ∶ 0.3]
 If the shape of the plot is wrong, in that case, the following would be used. If in the detailed work,
 it is shown that 𝑁 = 𝑚𝑔 for 𝜔 < 𝜔𝑐 , 0.1 points would be provided. If it is shown that 𝑁 = 𝑚𝜔2 𝑅 for
 𝜔 ≥ 𝜔𝑐 , 0.1 points would be provided.

Solutions

                                                                                A2-4
                                                                                Oﬃcial (English)

 B.4 (1.0 pt)
 The potential 𝑉 (𝜃)

 Solution 1: Using direct integration
 Given that
                                                   1 𝑑𝑉 (𝜃)
                                                       𝐹𝜃 = −                                          (7)
                                                   𝑅 𝑑𝜃
 and taking 𝑉 (𝜃 = 0) = 0, we obtain on integrating Eq. (4) that
                                                   𝜃                𝑉
                                           −𝑅 ∫ 𝐹𝜃 𝑑𝜃 = ∫ 𝑑𝑉 = 𝑉 − 0
                                               0                0

                                                                                [0.3]
 the left hand side is
                                𝜃                               𝜃                     𝜃
                                               −𝑚𝜔2 𝑅2
                         −𝑅 ∫ 𝐹𝜃 𝑑𝜃        =           ∫ sin(2𝜃) + 𝑚𝑔𝑅 ∫ sin(𝜃)𝑑𝜃
                            0                    2     0               0
                                               𝑚𝜔2 𝑅2 (cos(2𝜃) − 1)
                                           =                        − 𝑚𝑔𝑅(cos(𝜃) − 1)                  (8)
                                                        4
                                                                                [0.4]
                                       2
 Noting that 𝑐𝑜𝑠(2(𝜃) − 1) = −2 sin (𝜃) and 𝜔𝑐2 = 𝑔/𝑅 we obtain

                                                                         𝜔2     2
                                    𝑉 (𝜃) = 𝑚𝑔𝑅 [(1 − cos 𝜃) −               sin 𝜃]                    (9)
                                                                        2𝜔𝑐2

                                                                                [0.3]
 We can also verify the above equation by substitution into Eq. (7)
 𝑃 = 𝑚𝑔𝑅
 𝑄 = −𝑚𝑔𝑅
       𝜔2 𝑚𝑔𝑅
 𝑆=−
         2𝜔𝑐2
                                                     2
 Solution 2: Differentiating 𝑉 = 𝑃 + 𝑄 cos(𝜃) + 𝑆 sin (𝜃)
                                      1 𝑑𝑉 (𝜃)   𝑄          𝑆
 𝐹𝜃 = 𝑚𝑅 sin 𝜃 cos 𝜃𝜔2 − 𝑚𝑔 sin 𝜃 = −          = sin(𝜃) − 2 sin 𝜃 cos 𝜃
                                      𝑅 𝑑𝜃       𝑅          𝑅
                                                                                [0.3]
 Comparing, we get,
 𝑄 = −𝑚𝑔𝑅
      𝜔2 𝑚𝑔𝑅
 𝑆=−
        2𝜔𝑐2
                                                                                [0.2,0.2]
 Also, as 𝑉 (0) = 0, we have 𝑃 + 𝑄 = 0. Hence, 𝑃 = 𝑚𝑔𝑅
                                                                                [0.3]

Solutions

                                                                             A2-5     Oﬃcial (English)

 B.5 (1.0 pt)
 The coeﬃcients

 We use the expansions for the trigonmetric functions sin(𝜃) and cos(𝜃) in Eq. (10). We shall keep
 terms upto and inculding order 𝜃4 . Thus

                                                            𝜔2
                       𝑉 (𝜃)   ≈   𝑚𝑔𝑅 [1 − 1 + 𝜃2 /2 − 𝜃4 /24 − (𝜃 − 𝜃3 /6)2 ]
                                                            2𝜔𝑐2
                                   𝑚𝑔𝑅     𝜔2        𝑚𝑔𝑅 𝜔2     1
                               ≈       [1 − 2 ] 𝜃2 +    [ 2 − ] 𝜃4
                                    2      𝜔𝑐         6  𝜔𝑐     4

 Thus
                                                  𝑚𝑔𝑅     𝜔2
                                         𝑎(𝜔) =       (1 − 2 )
                                                   2      𝜔𝑐
                                                                                [0.5]

                                                  𝑚𝑔𝑅 𝜔2  1
                                         𝑏(𝜔) =      (   − )
                                                   6 𝜔𝑐2  4
                                                                                [0.5]
 Note: no penalty if the 1/4 term is missed.
 One observes that if one incorrectly expands sin 𝜃 ≈ 𝜃, in that case, only 𝑎(𝜔) will turn out to be
 correct.

Solutions

                                                          A2-6
                                                           Oﬃcial (English)

 B.6 (1.0 pt)
 Representative plots of the potential

 Solution 1: Plotting for 𝜃 ∈ [−𝜋/2, 𝜋/2]
                                             𝜔 < 𝜔𝑐
                                             𝜔 > 𝜔𝑐

 𝑉 (𝜃)

         0

                −𝜃0                           𝜃0

         − 𝜋2                   0                     𝜋
                                                      2
                                𝜃
                                                             [𝜔 < 𝜔𝑐 ∶ 0.5]
                                                             [𝜔 > 𝜔𝑐 ∶ 0.5]
 Solution 2: Plotting for 𝜃 ∈ [−𝜋, 𝜋]
                                             𝜔 < 𝜔𝑐
                                             𝜔 > 𝜔𝑐

 𝑉 (𝜃)

         0
                       −𝜃0              𝜃0

         −𝜋                     0                     𝜋
                                𝜃
                                                             [𝜔 < 𝜔𝑐 ∶ 0.5]
                                                             [𝜔 > 𝜔𝑐 ∶ 0.5]

Solutions

                                                                                A2-7      Oﬃcial (English)

  B.7 (1.0 pt)
 Bead analogues
 Solution 1:
 For 𝜔 → 𝜔𝑐+ , 𝜃0 is close to zero. Hence on expanding the cosine term in Eq. (5),

                                            𝜃02        𝜔𝑐2
                                      1−           =
                                            2          𝜔2
                                                                    1/2
                                                         √     𝜔2
                                            𝜃0     =   ± 2 [1 − 𝑐2 ]                                   (10)
                                                               𝜔

 Also note from Eq. (5) that as 𝜔 → ∞, 𝜃0 → ±𝜋/2. This plot also has an analogue in phase transition.
 The magnetization ℳ goes to zero as 𝑇 goes to 𝑇𝑐 in a similar fashion. Thus the role of ℳ is played
 by 𝜃0 and temperature is inversely related to 𝜔. Increasing temperature is equivalent to decreasing
 𝜔. Summarizing,
                                               ℳ⟶𝜃
                                                                                [0.4]

                                                   𝑇𝑐 ⟶ 1/𝜔𝑐2
                                                  𝑇 /𝑇𝑐 ⟶ 𝜔𝑐2 /𝜔2
                                                                                [0.4]
 Equivalent value of 𝛽 for bead is = 1/2.
                                                                                [0.2]
 Solution 2:
                                                      2
 For 𝜔 > 𝜔𝑐 , cos 𝜃0 = 𝜔𝑐2 /𝜔2 . Hence on writing sin 𝜃0 = 1−cos2 𝜃0 and substituting the value of cos 𝜃0 ,
                        4
 one gets sin 𝜃0 = (1− 𝜔𝑐 1/2
                       𝜔4 )   . This plot also has an analogue in phase transition. The magnetization ℳ
 goes to zero as 𝑇 goes to 𝑇𝑐 in a similar fashion. Thus the role of ℳ is played by sin 𝜃0 (or equivalently
 𝜃0 in the small angle limit) and temperature is inversely related to 𝜔4 . Increasing temperature is
 equivalent to decreasing 𝜔. Summarizing,

                                                   ℳ ⟶ sin 𝜃

                                                                                [0.4]

                                                   𝑇𝑐 ⟶ 1/𝜔𝑐4
                                                  𝑇 /𝑇𝑐 ⟶ 𝜔𝑐4 /𝜔4
                                                                                [0.4]
 Equivalent value of 𝛽 for bead is = 1/2.
                                                                                [0.2]
 [Note: The critical exponent is 1/2 in our case and also in Landau theory. However experimentally
 and in more elaborate theories the exponent of vanishing magnetization is 1/3].

Solutions

                                                                                A2-8
                                                                                Oﬃcial (English)

 B.8 (1.0 pt)
 Oscillation frequency

 The frequency of oscillation Ω0 of the bead about the "equilibrium" position 𝜃0 is

                                                        1 √ 𝑉 ″ (𝜃)
                                                 Ω0 =
                                                        𝑅     𝑚
 We take the second order derivative of the potential as given in Eq. (10)

                                                   𝜔2             𝜔2    2
                       𝑉 ″ (𝜃) = 𝑚𝑔𝑅 cos 𝜃 [1 −     2
                                                      cos 𝜃] + 𝑚𝑔𝑅 2 sin 𝜃                          (11)
                                                   𝜔𝑐             𝜔𝑐

 For 𝜃 = 𝜃0 = ± cos−1 (𝜔𝑐2 /𝜔2 )

                                            𝜔2       𝜔𝑐4
                          𝑉 ″ (𝜃0 ) = 𝑚𝑔𝑅       (1 −     )>0                 if 𝜔 > 𝜔𝑐              (12)
                                            𝜔𝑐2      𝜔4

 For 𝜔 < 𝜔𝑐 , 𝜃0 = 0, and we obtain from Eq. (12) that

                                                 Ω0 = (𝜔𝑐2 − 𝜔2 )1/2                                (13)

                                                                                [0.5]
 Similarly for 𝜔 > 𝜔𝑐 , using Eq. (13) we obtain
                                                                       1/2
                                                              𝜔4
                                            Ω0     =    𝜔 (1 − 𝑐4 )                                 (14)
                                                              𝜔

                                                                                [0.5]
 No credit will be provided is small angle approximation of 𝑉 (𝜃) is used.

Solutions

                                                                           A2-9      Oﬃcial (English)

 B.9 (1.0 pt)
 Sketch of Ω0

                                           𝜔 < 𝜔𝑐
                                           𝜔 > 𝜔𝑐

                              𝜔
 Ω0                       0 =
                          Ω

       𝜔𝑐

                     𝜔𝑐
                                𝜔
                                                                                [𝜔 < 𝜔𝑐 ∶ 0.5]
                                                                                [𝜔 > 𝜔𝑐 ∶ 0.5]
 In the case of wrong expression of Ω0 derived in the previous part, marks would be awarded based
 on the plot of expression obtained, and physicality of the plots.

Solutions

                                                                                A2-10          Oﬃcial (English)

 C.1 (1.0 pt)
 Condition for equilibrium angles

 We substitute the expression for the normal reaction (Eq.(1)) in the angular part (Eq.(3)) to obtain
                                                                             2
           𝑚𝑅𝜃 ̈ =     𝑚𝑅 sin(𝜃) cos(𝜃)𝜔2 − 𝑚𝑔 sin(𝜃) − 𝑓𝑘(𝑚𝑔 cos(𝜃) + 𝑚𝑅 sin (𝜃)𝜔2 + 𝑚𝑅𝜃 ̇2 )

 Noting that 𝜔𝑐2 = 𝑔/𝑅 and rearranging terms we have

                                                         2                             2
                    2⎡                                 𝜔                            𝜃̇
           𝜃̈ =    𝜔𝑐 ⎢(sin(𝜃)) (cos(𝜃) − 𝑓𝑘 sin(𝜃)) ( ) − sin(𝜃) − 𝑓𝑘 cos(𝜃) − 𝑓𝑘 ( ) ⎤ ⎥
                                                      𝜔𝑐                            𝜔𝑐
                       ⎣                                                                           ⎦
                                                                                [0.2]
 At equilibrium, 𝜃 ̇ = 0, 𝜃 ̈ = 0 and 𝑓 = sgn(𝜃)̇ = ±1 depending on how this equilibrium was attained,
 iė., depending on the value of 𝜃 ̇ just before equilibrium was attained. Thus we obtain the expression
 for the equilibrium angle 𝜃0 ,

                                                      2
                                                  𝜔
            sin(𝜃0 ) (cos(𝜃0 ) − 𝑓𝑘 sin(𝜃0 )) (      ) = sin(𝜃0 ) + 𝑓𝑘 cos(𝜃0 ) with 𝜃0 ∈ (−𝜋/2, 𝜋/2)
                                                  𝜔𝑐

                                                                                [0.4]
 For 𝑓 = 1 and 𝑘 = tan(𝛼) we may express the above as
                                      2
                                𝜔                      sin(𝜃0 ) + tan(𝛼)𝑐𝑜𝑠(𝜃0 )
                               ( )        =
                                𝜔𝑐                sin(𝜃0 )(cos(𝜃0 ) − tan(𝛼) sin(𝜃0 ))
                                                  tan(𝜃0 + 𝛼)
                                          =                                                                (15)
                                                    sin(𝜃0 )

                                                                                [0.4]
 In case of algebraic error leading to 𝑥 = 𝜃0 − 𝛼, only 0.1 points would be deducted.

Solutions

                                                                    A2-11            Oﬃcial (English)

 C.2 (0.5 pt)
 Representative values for 𝜃0
 We are given the expansions for the trignometric functions in the problem. We notice that the coef-
 ﬁcient of the opposing force 𝑘 is small (=0.05). Thus 𝑘 = 𝛼. We then have

                                              sin(𝜃0 ) ≈ 𝜃0

                                         tan(𝜃0 + 𝛼) ≈ 𝜃0 + 𝛼
                                                                                [0.2]
 Thus
                                                    2
                                             𝜔                 𝑘
                                         (      )       ≈ 1+
                                             𝜔𝑐                𝜃0

 Simple calculations yield
 (a) 𝜃0 = -0.07 radians
 (b) 𝜃0 = -0.1 radians

                                                                                [0.3]
 The plot will no longer be symmetric.

APhO 2022 Theory T2


[User]
2. COMET (8 points)

— Jaan Kalda. A
comet’s orbit intersects the Earth’s orbit (which
can be assumed to be circular of radius R0 =
1.5×108
km) at an angle α = 45°. The comet’s
and the Earth’s orbits lie on the same plane.
i)(3points)FindthedistanceRmin ofthecomet’s
perihelion P from the Sun (i.e. its shortest
distance). The distance Rmax of the comet’s
aphelion A from the Sun (i.e. its longest dis-
tance) can be assumed to be much larger than
R0.
ii) (5 points) For how many days t will the
comet’s distance to the Sun be less that R0?

Solution:
2. COMET

i) Sincethedistancetoaphelionisverylarge,the
comet’s full energy −G M⊙
Rmin+Rmax
can be taken
to be zero and near the Sun the orbit is approx-
imately shaped like a parabola. Hence, at the
distance R0, 1
2v2
= G M⊙
R0
(the same result can
be found by writing out the conservation of an-
gular momentum and energy for the comet at
two points: the aphelion and the intersection of
the two orbits; and by using Rmax ≫ R0). Since
α = 45°, the tangential and radial velocities are
equal, vt = vr and v2
≡ v2
t +v2
r =2v2
t , therefore
v2
t =G
M⊙
R0
(which is exactly the same expression what we
have for the orbital speed w of Earth). Angu-
lar momentum conservation law allows us to ex-
press the speed at perihelion u = vtR0/Rmin;
from energy conservation law 1
2 u2
= G M⊙
Rmin
.
Thus,
1
2
GM⊙
R0
·
R2
0
R2
min
=
GM⊙
Rmin
from where Rmin = 1
2R0.
ii) Denote by O the focus of the elliptical or-
bit of the comet where the Sun is, and by Q the
other focus of the comet’s orbit. Let B and C
be the intersection points of the Earth’s and the
comet’s orbits.
B
O
C
Q
P
We will show that the points B, C and O
are all approximately on the same line by prov-
ing that the angle ̸ QOB is approximately 90°.
By the property of the ellipse that the sum of
thelengths of theline segments drawnfromany
pointontheellipsetothetwofocalpointsiscon-
stant we get
|QB|+|BO|=|QP|+|OP|=|QO|+2|OP|.
By using |BO| = R0 and that approximately
|OP| = 1
2R0 (the equality is in the limit that
Rmax goes to infinity), we get that |QB| =
|QO|. So the triangle △QBO is isosceles and
theangles̸ QOB and̸ QBO areequal. Because
|QB| ≫ |OB|, we have approximately ̸ QOB =
̸ QBO = 90° and all the points B, C and O are
approximately on the same line.
Theradiusvectorofthecometcoversanarea
SBCP contained between the segment BCP of
the ellipse and the straight line BC. We will ap-
proximate this segment of the ellipse as a para-
bola. The surface area of a mirror-symmetric
parabolic segment is equal to two thirds of the
product of its height and length, as can be de-
termined by integrating the parabola:
SBCP =
2
3
·2R0 ·
R0
2
=
2
3
R2
0.
ByKepler’ssecondlawthecometsweepsout
equal areas in equal times (i.e. the time required
to cover some segment of the ellipse is propor-
tional to the area of the segment drawn out by
the radius vector connecting the comet and the
Sun). Since vt = w, the radius vector of the
comet covers the surface area at the same rate as
the radius vector of the Earth, which is πR2
0/T,
where T is one year. Therefore
t=
SBCP
πR2
0
T =
2
3π
365days≈77days.
Alternative solution: We can also prove that
the intersection points of the orbits and the sun
lie on the same line in the following way. Near
the Sun we can approximate the orbit as a para-
bola. Due to the geometrical property of a para-
bola, if we take the x-axis parallel to be the axis
of the parabola pointing towards the perihelion
P, and the origin O at the focus (i.e. the Sun)
then for any point S at the parabola, |OS|+x=
2|OP| = R0, where x is the x-coordinate of
S. For the intersection points B and C of the
Earth’s and comet’s orbits. |OB| = |OC| = R0,
therefore for the both points x = 0, i.e. O, A,
and B all lie on the y-axis.
Alternative solution: Here is an alternative
solution to show that the points B, C and O are
all approximately on the same line. We have
|OQ|= Rmax −Rmin = Rmax −
R0
2
.
Every ellipse has the property that the lines
drawn from any point on the ellipse to the two
focal points are at the same angle to the tan-
gent. Using this property, because the tangent
at point B is at 45° to the segment OB, then
|BQ| is also at 45° to the tangent and ̸ OBQ =
90°. So in the right angled triangle △BOQ we
have
cos̸ BOQ =
|OB|
|OQ|
=
R0
Rmax −
R0
2
≈0,
where we used the fact that Rmax ≫ R0. So we
have approximately ̸ BOQ =90° and the points
B, C and O all lie on the same line.

COMET


[User]
01@
                      Do the following JPhO problem. This pedagogical problem introduces the WKB
    approximation and phase space, reviewing everything covered above, and applies it to “clusters” of
    atoms. You can skip sections I and III, which are covered elsewhere on this problem set.

        Remark
        Because matter has wave properties, particles such as electrons can exhibit interference effects,
        like those covered in W2. For example, you can run a double-slit experiment firing individual
        electrons at a time and seeing where they hit the screen, which will gradually build up an
        interference pattern. You’ll see an example of this in the next problem.

Solution:
See the official solutions here.

        Remark
        Because matter has wave properties, particles such as electrons can exhibit interference effects,
        like those covered in W2. For example, you can run a double-slit experiment firing individual
        electrons at a time and seeing where they hit the screen, which will gradually build up an
        interference pattern. You’ll see an example of this in the next problem.