[User]
Consider two concentric rings of radii r and R ≫ r.
     (a) Compute the mutual inductance by considering a current through the larger ring.
     (b) Compute the mutual inductance by considering a current through the smaller ring, and verify
         your results agree. (Hint: this can be done without difficult integrals.)
    In general, computing mutual inductance is a hard and practically important problem; there have
    been whole books written on the subject.
[2] Problem 20 (MPPP 181). Three nearly complete circular loops, with radii R, 2R, and 4R are
    placed concentrically on a horizontal table, as shown.

    A time-varying electric current is made to flow in the middle loop. Find the voltage induced in the
    largest loop at the moment when the voltage between the terminals of the smallest loop is V0 .

    4    Magnetism
    In this section we’ll dip a little into atomic physics and the origin of magnetism. However, a proper
    understanding of this subject requires quantum mechanics, as we’ll cover in T3 and X3.

  Idea 6
  A spinning object of charge q and mass m carries a magnetic dipole moment µ and angular
  momentum J. If its mass and charge distributions are proportional, then µ and J point in
  the same direction, and in classical mechanics, their ratio is always µ/J = q/2m.

  In quantum mechanics, it turns out that µ/J = qg/2m, where the g-factor is an order-one
  number, which is about 2 for the electron. Unfortunately, there’s no way to derive this result
  with the Olympiad syllabus, so problems about magnetism can at best make rough estimates.

  Example 7

  Suppose the magnetic moment of an iron atom is due to a single unpaired electron, with
  angular momentum of order ℏ. The atoms are separated by a distance d ∼ 10−10 m. Estimate
  the maximum magnetic field an iron magnet can produce. How does this compare to the
  fields that can be produced in an electromagnet?

  Solution
  The answer doesn’t scale significantly with the physical size of the iron magnet. To see this,
  think in terms of electric dipoles: if you have a giant cube of electric dipoles, it’s equivalent
  to having a fixed surface charge density ±σ on two of the faces. The electric field produced
  by such a charge density near each face is of order σ/ϵ0 , independent of the size of the cube.

  Therefore, the only things the magnetic field can depend on are µ0 , the magnetic dipole
  moment µ of a single atom, and d. By dimensional analysis,
                                                  µ
                                         B ∼ µ0 3
                                                 d
  which can also be thought of as µ0 M , where M is the magnetization density. Taking
  µ ∼ eℏ/me and plugging in the numbers gives B ∼ 10 T, which is the right order of magnitude.

  Now consider the case of an electromagnet, where the field is produced by moving electrons
  with typical speed v, moving in a loop with typical size r. In a metal, there’s on the order
  of one free electron per atom, so d is still the same. The difference is that the field made by
  each electron does scale with r, because each has magnetic moment
                                                     ev 2
                                          µ = IA ∼      r .
                                                      r
  Compared to the previous result, this is larger by a factor of mvr/ℏ. The two are comparable,
  for r ∼ 1 m, if the electrons travel at the agonizingly slow velocity v ∼ 10−4 m/s.

  Therefore, you would get a magnetic field much larger than 10 T if you could make the
  electrons go at a reasonable walking speed, but that’s easier said than done. The largest
  steady magnetic fields made in the lab are only about 40 T. Such a field carries a pressure
  which would rip apart a solenoid made of coiled wire,
                                           (40 T)2
                                     P =           = 0.6 GPa.
                                             2µ0

         Instead, these fields are produced in Bitter electromagnets, which are solenoids made of thick
         metal plates, perforated with cooling channels to dissipate the enormous heat produced by
         resistance. It is possible to produce higher fields temporarily, but the results will be explosive.

Solution:
(a) The field at the center is µ0 I/2R, so the flux through the small ring is Φ =
                                   πr2
        (µ0 I/2R)πr2 , so L12 = µ02R   .

     (b) Consider the entire infinite plane the smaller ring lies in. The key idea is that the total flux
         through this plane is zero: every magnetic flux line due to the ring that goes up through the
         plane comes down through it somewhere else. Now decompose this plane into the part in the
         big ring and the part outside. We have

                                             Φplane = Φin + Φout = 0

         which means Φin = −Φout . This is useful because calculating Φin is very complicated. Cal-
         culating Φout is easy because the whole region is far from the small ring, so its field can be
         approximated as a dipole field.

         The field at a distance s away in the plane of the rings is B = Bz ẑ where Bz = µ4πs
                                                                                0 Iπr
                                                                                3 . The

         flux is then
                                                 µ0 Ir2 ∞ 1              µ0 Iπr2
                                                        Z
                                Φin = −Φout = −               3
                                                                2πs ds =
                                                   4     R s               2R
         Dividing by I gives the same result as (a), as expected.

[2] Problem 20 (MPPP 181). Three nearly complete circular loops, with radii R, 2R, and 4R are
    placed concentrically on a horizontal table, as shown.

   A time-varying electric current is made to flow in the middle loop. Find the voltage induced in the
   largest loop at the moment when the voltage between the terminals of the smallest loop is V0 .
   Solution. As you might have noticed from the previous problems, inductances generally scale with
   one power of length. In this case, the only length scale is the radii of the rings, so the mutual
   inductance of the 2R and 4R loops is twice that of the 2R and R loops by dimensional analysis.
   Since V0 = LR,2R I˙2R , we have V4R = L2R,4R I˙2R = 2V0 .
       But there’s a subtlety: in problem 18, we found that the self-inductance of a wire loop depended
   on the radius r of the wire, so why didn’t we allow it in the dimensional analysis here? The point
   is that the self-inductance of a wire loop depends on the flux the loop puts through itself, so it
   depends on the very large magnetic fields very closer to the wire. But a mutual inductance only
   depends on the flux one loop puts through another, and here, all points of the second loop are far
   away from the first loop. So the wire radius doesn’t matter, just like how it didn’t in problem 19.

   4    Magnetism
   In this section we’ll dip a little into atomic physics and the origin of magnetism. However, a proper
   understanding of this subject requires quantum mechanics, as we’ll cover in T3 and X3.

       Idea 6
       A spinning object of charge q and mass m carries a magnetic dipole moment µ and angular
       momentum J. If its mass and charge distributions are proportional, then µ and J point in
       the same direction, and in classical mechanics, their ratio is always µ/J = q/2m.

       In quantum mechanics, it turns out that µ/J = qg/2m, where the g-factor is an order-one
       number, which is about 2 for the electron. Unfortunately, there’s no way to derive this result
       with the Olympiad syllabus, so problems about magnetism can at best make rough estimates.

       Example 7

       Suppose the magnetic moment of an iron atom is due to a single unpaired electron, with
       angular momentum of order ℏ. The atoms are separated by a distance d ∼ 10−10 m. Estimate
       the maximum magnetic field an iron magnet can produce. How does this compare to the
       fields that can be produced in an electromagnet?

       Solution
       The answer doesn’t scale significantly with the physical size of the iron magnet. To see this,
       think in terms of electric dipoles: if you have a giant cube of electric dipoles, it’s equivalent
       to having a fixed surface charge density ±σ on two of the faces. The electric field produced
       by such a charge density near each face is of order σ/ϵ0 , independent of the size of the cube.

       Therefore, the only things the magnetic field can depend on are µ0 , the magnetic dipole
       moment µ of a single atom, and d. By dimensional analysis,
                                                           µ
                                                  B ∼ µ0
                                                           d3
       which can also be thought of as µ0 M , where M is the magnetization density. Taking
       µ ∼ eℏ/me and plugging in the numbers gives B ∼ 10 T, which is the right order of magnitude.

       Now consider the case of an electromagnet, where the field is produced by moving electrons
       with typical speed v, moving in a loop with typical size r. In a metal, there’s on the order
       of one free electron per atom, so d is still the same. The difference is that the field made by
       each electron does scale with r, because each has magnetic moment
                                                           ev 2
                                               µ = IA ∼       r .
                                                            r
       Compared to the previous result, this is larger by a factor of mvr/ℏ. The two are comparable,
       for r ∼ 1 m, if the electrons travel at the agonizingly slow velocity v ∼ 10−4 m/s.

       Therefore, you would get a magnetic field much larger than 10 T if you could make the
       electrons go at a reasonable walking speed, but that’s easier said than done. The largest
       steady magnetic fields made in the lab are only about 40 T. Such a field carries a pressure
       which would rip apart a solenoid made of coiled wire,

                                                 (40 T)2
                                           P =           = 0.6 GPa.
                                                   2µ0
       Instead, these fields are produced in Bitter electromagnets, which are solenoids made of thick
       metal plates, perforated with cooling channels to dissipate the enormous heat produced by
       resistance. It is possible to produce higher fields temporarily, but the results will be explosive.


[User]
Matkaja rõõmustab laagriplatsil, et elektrijaama müra tuuletu ilmaga nii vaikselt temani kostab. Hiljem, kui tuul tõuseb, on müra veel vaiksem. Puhub põhjatuul kiirusega $\beta c$, kus $c$ on heli kiirus paigalseisvas õhus; jaam jääb matkajast edelasse
(st tuule ja jaama suundade vaheline nurk on $\alpha = \ang{135}$).\\
\osa Kas helisagedus on sama mis tuuleta?\\
\osa Kui tuuleta on tajutav helivõimsus $P$ ja tuulega $xP$, siis kui suur on $x$?\\
Võite lugeda, et elektrijaam on punktikujuline.

\emph{Soovitus:} uurige helifrondi levimist.

Solution:
% Joonis
Õhu taustsüsteemis on helilainefrondid kiirusega $c$ kasvava raadiuse ja paigaloleva keskmega poolsfäärid. Helivõimsus jaotub ühtlaselt üle terve frondipinna, seega on tajutav võimsus pöördvõrdeline frondi pindalaga ehk pöördvõrdeline frondi raadiuse ruuduga. Lisaks on võimsus ka võrdeline frontide vastuvõtmise sagedusega ehk võrdeline helisagedusega, aga nagu peagi leiame, on sagedus konstant.

\begin{wrapfigure}{r}{0.5\linewidth}
	\vspace{-10pt}
	\begin{center}
		\includegraphics[width = 0.9\linewidth]{2009-lahg-10-lah}
	\end{center}
\end{wrapfigure}

Õhu taustsüsteemis liiguvad nii matkaja kui ka jaam vastu esialgsele tuule suunale kiirusega $\beta c$. Järelikult kui front oli vastuvõtmise hetkeks raadiusega $d = ct$, siis selle kiirgamise alguspunkt oli liikunud jaama suhtes allatuult kaugusele $a = (\beta c)t = \beta d$. Et see kaugus on kõigile frontidele ühesugune, on sama ka frontide teeloleku aeg ning aeg kahe frondi kiirgamise vahel võrdub ajaga nende vastuvõtmise vahel. Seega helisagedus ei muutu. Olgu jaama kaugus $\ell$. Rakendame tekkinud kolmnurgale koosinusteoreemi:
\[
d^{2}=a^{2}+\ell^{2}-2 a \ell \cos \alpha.
\]
Kuna $x = \left(\frac{\ell}{d}\right)^2$ ja $\cos \ang{135}=-\frac{\sqrt{2}}{2}$, saame $\sqrt x$ leidmiseks ruutvõrrandi, kusjuures $\ell$ taandub välja (karakteristlik pikkusmõõde puudub).
\[
\begin{array}{c}{\frac{\ell^{2}}{x}=\frac{\beta^{2} \ell^{2}}{x}+\ell^{2}+\frac{\sqrt{2} \beta \ell^{2}}{\sqrt{x}}} \\ {x+\sqrt{2} \beta \sqrt{x}+\beta^{2}-1=0} \\ {x=\left(\frac{-\sqrt{2} \beta \pm \sqrt{2 \beta^{2}-4 \beta^{2}+4}}{2}\right)^{2}=\frac{\left(-\beta \pm \sqrt{2-\beta^{2}}\right)^{2}}{2}=1 \mp \beta \sqrt{2-\beta^{2}}.}\end{array}
\]
Heli jääb vaiksemaks, mistõttu $x < 1$ ja peame valima miinusmärgiga lahend
\[
x=1-\beta \sqrt{2-\beta^{2}}.
\]

Müra

Matkaja rõõmustab laagriplatsil, et elektrijaama müra tuuletu ilmaga nii vaikselt temani kostab. Hiljem, kui tuul tõuseb, on müra veel vaiksem. Puhub põhjatuul kiirusega $\beta c$, kus $c$ on heli kiirus paigalseisvas õhus; jaam jääb matkajast edelasse
(st tuule ja jaama suundade vaheline nurk on $\alpha = \ang{135}$).\\
\osa Kas helisagedus on sama mis tuuleta?\\
\osa Kui tuuleta on tajutav helivõimsus $P$ ja tuulega $xP$, siis kui suur on $x$?\\
Võite lugeda, et elektrijaam on punktikujuline.

\emph{Soovitus:} uurige helifrondi levimist.

Ülesannet on mugavam vaadelda õhu taustsüsteemis, sest siis on helilainefrondid kiirusega $c$ kasvava raadiusega ja paigaloleva keskmega poolsfäärid. Helivõimsus jaotub ühtlaselt üle terve frondipinna, seega on tajutav võimsus pöördvõrdeline frondi pindalaga ehk frondi raadiuse ruuduga.


[User]
One of the important achievements of the 19th century was the laying of undersea
    cables, which permitted the transmission of telegraph messages. In 1871, the 21 year old electrician
    Oliver Heaviside was tasked with locating a leak in the cable connecting England and Denmark.
    (Heaviside had been trained by his uncle-in-law Charles Wheatstone, who found many uses for the
    Wheatstone bridge. Heaviside later recast Maxwell’s equations in the vector form we use today.)
        The cable can be modeled as a uniform cylinder of known resistance R0 . That is, when the cable
    is operating properly, then grounding one end and applying a voltage V to the other leads to a
    steady state current of V /R0 . The leak is located a fraction α of the way from the English side. Let
    the resistance between the leak point and the Earth, due to the current having to travel through
    the water, be Rd . The precise value of Rd is also unknown. Your task, as was Heaviside’s, is to find

                  01W
    a way to measure α without having to dig the whole cable up.

Solution:
There are various ways to solve this problem; here’s what Heaviside did.

    First, don’t attach the Danish side to anything, and apply a voltage at the English side. By
    measuring the resulting current, we can measure the resistance, which in this case is

                                              R1 = αR0 + Rd .

    Next, attach the Danish side to ground and repeat the procedure to measure the new resistance,
                                                       Rd R0 (1 − α)
                                       R2 = αR0 +                     .
                                                      Rd + R0 (1 − α)
    We don’t know Rd , but we can plug the first equation into the second to eliminate it. Defining the
    rescaled variables r1 = R1 /R0 and r2 = R2 /R0 , the second equation becomes

                                                    (r1 − α)(1 − α)
                                         r2 = α +                   .
                                                      r1 + 1 − 2α
    Clearing denominators and simplifying gives the quadratic

                                       α2 − 2αr2 + r2 + r1 r2 − r1 = 0

    which has solution                             p
                                        α = r2 ±    (r2 − r1 )(r2 − 1).
    Incidentally, this result was first derived by a French telegrapher, and is called Blavier’s method.
    Variations of this method are still used to locate breaks in cables today!

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


[User]
01mUSAPhO 2002, problem B1. This one is trickier than it looks! It can be solved
    with either a torque or energy analysis, but both require care.

[4] Problem 30 (IPhO 1982). A coat hanger can perform small oscillations in the plane of the figure
    about the three equilibrium figures shown.

    In the first two figures, the long side is horizontal. The other two sides have equal length. The
    period of oscillation is the same in all cases. The coat hanger does not necessarily have uniform
    density. Where is the center of mass, and how long is the period?

[4] Problem 31 (APhO 2007). A uniform ball of mass M and radius r is encased in a thin spherical
    shell, also of mass M . The shell is placed inside a fixed spherical bowl of radius R, and performs
    small oscillations about the bottom. Assume that friction between the bowl and shell is very large,
    so the shell essentially always rolls without slipping.
       The ball is made of an unusual material: it can quickly transition between a liquid and solid
    state. When the ball is in the liquid state, it has no viscosity, and hence no friction with the shell.
    When the ball is in the solid state, it rotates with the shell.

      (a) Find the period of the oscillations if the ball is always in the solid state.

      (b) Find the period of the oscillations if the ball is always in the liquid state.

      (c) The ball is now set so that it instantly switches to the liquid state whenever it starts moving
          downward, and instantly switches to the solid state whenever it starts moving upward. If the
          initial amplitude of oscillations is θ0 , find the amplitude after n oscillations.

Solution:
See the official solutions here.
[4] Problem 31 (APhO 2007). A uniform ball of mass M and radius r is encased in a thin spherical
    shell, also of mass M . The shell is placed inside a fixed spherical bowl of radius R, and performs
    small oscillations about the bottom. Assume that friction between the bowl and shell is very large,
    so the shell essentially always rolls without slipping.
       The ball is made of an unusual material: it can quickly transition between a liquid and solid
    state. When the ball is in the liquid state, it has no viscosity, and hence no friction with the shell.
    When the ball is in the solid state, it rotates with the shell.
      (a) Find the period of the oscillations if the ball is always in the solid state.
      (b) Find the period of the oscillations if the ball is always in the liquid state.
      (c) The ball is now set so that it instantly switches to the liquid state whenever it starts moving
          downward, and instantly switches to the solid state whenever it starts moving upward. If the
          initial amplitude of oscillations is θ0 , find the amplitude after n oscillations.
    Solution.     (a) In the solid state, the inside rotates with the shell, so the moment of inertia is
                                              2      2         16
                                         I = M r2 + M r2 = M r2 .
                                              5      3         15
          The rolling without slipping condition means v = ωr, so the total kinetic energy is
                                    1           1               8        23
                               K = (2M )v 2 + Iω 2 = M v 2 + M v 2 = M v 2 .
                                    2           2              15        15
          If the angle between the line between the centers of the bowl and ball and the vertical is θ,
          then v = (R − r)θ̇, and the potential energy is
                                   U = 2M g(R − r)(1 − cos θ) ≈ M g(R − r)θ2 .
          Since both the kinetic and potential energy are quadratic, this is simple harmonic motion. As
          we saw in M4, if we write the total energy as
                                                  1          1
                                               E = meff θ̇2 + keff θ2
                                                  2          2
          then the period of oscillations is
                                                                   s
                                                                       23(R − r)
                                                   r
                                                       meff
                                          T = 2π            = 2π                 .
                                                       keff               15g

 (b) The only difference here is that the liquid will no longer rotate, so the first term in the moment
     of inertia above will no longer contribute. Then the kinetic energy is
                                           23       1      4
                                     K=       M v2 − M v2 = M v2
                                           15       5      3
     which implies, by the same logic as in part (a), that
                                                 s
                                                    4(R − r)
                                          T = 2π             .
                                                        3g

 (c) When the ball goes from solid to liquid, the entire ball is at rest, so no energy is lost. On
     the other hand, when the ball switches from liquid to solid, the material inside the ball must
     suddenly start rotating with the shell. This is an angular inelastic collision, where energy is
     lost, so we expect the amplitude to decay.
     Let’s suppose that just before the ball switches from liquid to solid, it has an angular velocity
     ωi . Then the total energy is

                                              Ei = M r2 ω02

     by the work we did in part (b). As the material solidifies, the angular momentum about the
     ball’s contact point with the ground is conserved. Let the final angular velocity be ωf .
     The initial moment of inertia of the shell about the contact point is
                                            2                5
                                       Ii = M r2 + M r2 = M r2 .
                                            3                3
     The shell is instantaneously rotating about the contact point, and so contributes angular mo-
     mentum Ii ωi . The liquid is only in translational motion, so it contributes angular momentum
     M vi r = M ωi r2 . Thus, the total angular momentum is

                                   Li = Ii ωi + M ωi r2 = M r2 ωi .

     After the transition, both the shell and ball will rotate about the contact point, and the
     moment of inertia is
                                     5        2                46
                               If = M r 2 + M r 2 + M r 2 = M r 2 .
                                     3        5                15
     Conserving the angular momentum gives
                                           8          46
                                             M r2 ωi = M r2 ωf
                                           3          15
     and therefore

                                                ωf =      ωi .

     Since the energy is E = ωL/2, this means

                                                Ef =      Ei .

                                 √                                                  p
     The angular amplitude θ ∝ E, so the amplitude decreases by a factor of 20/23 after each
     collision. But there are two collisions per oscillation, so after n oscillations, the amplitude is
                                                       n

                                             θ n = θ0         .


[User]
A block of mass m is bouncing back and forth in a box spanning 0 < x < L, with
    initial speed v0 . At time t = 0, the potential energy is slowly raised in part of the box, so that
                                                     (
                                                      V0 0 < x < ut
                                          V (x, t) =
                                                      0   ut < x < L

    where V0 > mv02 /2, and the speed of the potential u is very small. At time t = L/u, when the
    potential covers the entire box, what is the block’s speed?

[4] Problem 28 (F = ma, BAUPC). Two particles of mass m are connected by pulleys as shown.

    The mass on the left is given a small horizontal velocity v, and oscillates back and forth.

      (a) Without doing any calculation, which mass is higher after a long time?

     (b) Verify your answer is right by computing the average tension in the leftward string, in the
         case where the other end of the string is fixed, for amplitude θ0 ≪ 1.

      (c) Let the masses begin a distance L from the pulleys. Find the speed of the mass which
          eventually hits the pulley, at the moment it does, in terms of L and the initial amplitude θ0 .

Solution:
This is a simplified version of Physics Cup 2021, problem 4. The key is to use the
    adiabatic theorem. Initially, the box’s orbit in phase space is a rectangle with width ∆x = L and
    ∆px = 2mv0 . As the potential barrier enters the box, it effectively makes it shorter. So, just as in
    problem 26, the rectangle gets narrower while keeping its area the same.
       The twist is that eventually, the block will gain enough energy to climb over the potential barrier;
    at this point, the form of its orbit changes discontinuously, so we have to track exactly what’s going
    on instead of blindly using the adiabatic theorem. Consider the moment when the block’s energy
    becomes just enough to climb the potential barrier, and suppose that at this point, ut = x0 . Then
    the phase space orbit becomes the union of two rectangles:

       1. The original, shrinking rectangle with width ∆x = L − x0 and height ∆px = 2mv0 L/(L − x0 ).

       2. A new rectangle with width ∆x = x0 and negligible height.

    The added rectangle has negligible area, so the adiabatic invariant (the total phase space area)
    doesn’t change! After this point, we can continue to apply the adiabatic theorem until the end
    of the process. The first rectangle shrinks, until it reaches zero width, while the second rectangle
    grows. At the end of the process, we are back to a single rectangle with width ∆x = L and the
    same area as before, so the block ends up with the same speed as before.
       In terms of Newton’s laws, what’s going on is that the moving potential barrier is initially like
    a piston that does work on the block during each collision, but it also subtracts energy since the
    block has to eventually climb on top of it. Evidently, these two effects perfectly cancel, thanks to
    the adiabatic theorem.

[4] Problem 28 (F = ma, BAUPC). Two particles of mass m are connected by pulleys as shown.

    The mass on the left is given a small horizontal velocity v, and oscillates back and forth.

      (a) Without doing any calculation, which mass is higher after a long time?

     (b) Verify your answer is right by computing the average tension in the leftward string, in the
         case where the other end of the string is fixed, for amplitude θ0 ≪ 1.

      (c) Let the masses begin a distance L from the pulleys. Find the speed of the mass which
          eventually hits the pulley, at the moment it does, in terms of L and the initial amplitude θ0 .

Solution. (a) The mass on the right will be higher. If the masses didn’t move up or down, both
     would have the same average y-component of tension. But the mass on the left also has
     an x-component of tension, so its average magnitude of tension would be higher. This is a
     contradiction; to make the tension constant throughout the rope the mass on the right must
     rise.

 (b) The tension provides the centripetal force, so

                                                 mv 2
                                          T =         + mg cos θ
                                                  r
     where θ is the angle from the vertical. By energy conservation, the first term is 2mg(cos θ −
     cos θ0 ) where θ0 is the amplitude, so

                                       T = (3 cos θ − 2 cos θ0 )mg.

     Since the amplitude is small, we Taylor expand for

                                                 2   3 2
                                       T = 1 + θ0 − θ mg.

     Again using the small angle approximation, so that the motion is approximately simple
     harmonic, the average value of θ2 is θ02 /2, so

                                                  2  3 2
                                   T = 1 + θ0 − θ0 mg > mg

     as expected.

 (c) As we’ve seen above,
                                                        1 2
                                           T =       1 + θ mg

     where θ is the amplitude. Let x be the distance the right mass has risen. From the standpoint
     of the left mass, it is simply a pendulum whose length is being adiabatically lengthened, so
     by the result of problem 25, we have

                                        mg 2L 3/2 2 dx
                     Z                     Z
                                                                mg 2           1
                        (T − mg) dx =          L θ0 −3/2 =         Lθ0 1 − √        .
                                         4 L           x         2              2

     This is the increase in kinetic energy of the right mass, so setting this equal to mv 2 /2 gives
                                             s

                                         v = gL 1 − √          θ0 .


[User]
Derive the Carnot efficiency using the fact that (a) the engine is reversible, so a
    complete cycle leaves the entropy of the universe unchanged, or (b) directly from idea 1.

       Remark
       The most common mistake students make in this problem set is forgetting to account for the
       work done by the atmosphere.

Solution:
(a) At the end of a heat engine cycle, the working fluid comes back to its original state,
         so its entropy is unchanged. When an engine is reversible, it also leaves the entropy of the
         surroundings unchanged, so that the total entropy of the universe doesn’t go up. (If it did go
         up, then we wouldn’t be able to reverse the cycle, by the second law of thermodynamics.)
         The increase in entropy of the cold reservoir is Qout /TC , while the decrease in entropy of the
         hot reservoir is Qin /TH . Setting these equal to each other, we find
                                                 Qin − Qout     TC
                                            η=              =1−
                                                     Qin        TH
         as desired.

     (b) There are a lot of ways to do this, but we’ll show one that will be useful later. Note that
         during the heating and cooling steps, we have

                                       dQ = dU − dW = nCV dT + P dV.

         Using the ideal gas law, we have

                                            dT =      (P dV + V dP )
                                                   nR
         and plugging this in gives
                                                CP         CV
                                            dQ =   P dV +     V dP.
                                                R          R
         Dividing by T and using the ideal gas law again, we have
                                   dQ       dV       dP       d(P V γ )
                                      = nCP    + nCV    = nCV           .
                                    T       V        P         PV γ
         Therefore, the quantity ∆Q/T for the heating and cooling steps only depends on the change
         of the quantity P V γ . This is equal and opposite for those steps, since P V γ stays the same
         in the adiabatic steps, which means Qin /TH = Qout /TC . The derivation then continues as
         in part (a). (Of course, what we’ve done here is essentially just deriving the expression for
         the entropy, up to constants, without explicitly calling it that. This will be explored in more
         detail in T2.)

      Remark
      The most common mistake students make in this problem set is forgetting to account for the
      work done by the atmosphere.

                 01W
                 01W


[User]
Otsast kinnitatud varras saab pöörelda ümber horisontaaltelje ühes tasandis. Varda otsa on kinnitatud koormis massiga $m$. Varda pikkus on $l$. Varda kinnitusele mõjub hõõrdest tingitud pidurdav jõumoment $M$. Millistes nurkade vahemikes võib olla varras paigal (vt joonist)? Arvestada, et $mgl > M$.
\begin{center}
	\includegraphics[width=0.3\linewidth]{2008-v2g-01-yl}
\end{center}

Solution: Koormisele mõjub raskusjõu moment $mgl\sin \alpha$. Kang püsib paigal, kui see on väiksem hõõrdejõu momendist $M$, seega $mgl\sin \alpha < M$, millest $\sin \alpha < \frac{M}{mgl}$.

Pendel

Otsast kinnitatud varras saab pöörelda ümber horisontaaltelje ühes tasandis. Varda otsa on kinnitatud koormis massiga $m$. Varda pikkus on $l$. Varda kinnitusele mõjub hõõrdest tingitud pidurdav jõumoment $M$. Millistes nurkade vahemikes võib olla varras paigal (vt joonist)? Arvestada, et $mgl > M$.
\begin{center}
	\includegraphics[width=0.3\linewidth]{2008-v2g-01-yl}
\end{center}

Kangi kriitilise nurga korral kehtib varda jaoks jõumomentide tasakaal.


[User]
Quarterfinal 2005, problem 4. This is a neat example of separating out rotational
    and translational motion. For a similar idea, see Morin 8.73.

[2] Problem 10 (Morin 8.71). A ball sits at rest on a piece of paper on a table. You pull the paper
    in a straight line out from underneath the ball. You are free to pull the paper in an arbitrary way
    forward or backwards; you may even jerk it so that the ball starts to slip. After the ball comes off
    the paper, it will eventually roll without slipping. Show that, in fact, the ball ends up at rest. Is it
    possible to pull the paper in such a way that the ball ends up exactly where it started?

[2] Problem 11 (Morin 8.28). Consider the following “car” on an inclined plane.

    The system is released from rest, and there is no slipping between any surfaces. Find the acceleration

                    01W
    of the board.

                    01W

Solution:
The normal and gravitational forces cancel, so the only relevant force on the ball is
    friction, which acts at the bottom. Consider the angular momentum, L = r × p, and torques,
     τ = r × F, about the point of initial contact. RSince r and F are always in the same plane, τ always
    points perpendicular to the surface, and L = τ dt will likewise be vertical.
        During the process, the ball can move, as long as the horizontal components of its spin and
    orbital angular momentum cancel out. But after the ball comes off the paper, it will eventually roll
    without slipping, and in this case the spin and orbital angular momenta point in the same direction.
    So the only way for the sum to be zero is for both to be zero, so the ball stops.
        It is possible for the ball to end up where it started. If we just pull the paper out to the right,
    the ball ends up to the left of where it started. But we can do a little maneuver in the beginning
    to move the ball right, so that it cancels out the leftward motion in the last step. To do this, just
    jerk the paper to the right a bit, getting the ball started rolling to the right, then stop it later by
    jerking the paper to the left. Then pull the paper out to the right.

[2] Problem 11 (Morin 8.28). Consider the following “car” on an inclined plane.

    The system is released from rest, and there is no slipping between any surfaces. Find the acceleration
    of the board.

    Solution. Let the acceleration of the board be a, and the angular accelerations of the cylinders
    be α. Looking at one cylinder, the motion of the cylinder can be seen as pure rotation about the
    contact point with the slope (since there’s no slipping, that point is stationary). Then the cylinder

   rotates about the contact point with angular acceleration α, and the top will accelerate at α(2R)
   where R is the radius of the cylinders. Thus for the board to not slip, a = 2Rα.
      Taking torques about the contact point, with f being the friction force between the cylinders
   and board,
                                    m 2 1m 2             m
                              τ=       R +      R α = gR sin θ − 2Rf.
                                     2      2 2           2
   For the acceleration of the board,

                                           F = ma = 2f + mg sin θ.

   Adding these two equations and substituting αR = a/2 yields the answer,

                                                a=      g sin θ.

      This problem can also be solved using the “Lagrangian”/energy methods of M4. Let s be the
   distance the centers of the wheels have moved. Then by totaling up the kinetic energy,

                                       1 2         1         1
                                 K = mṡ × 1 + + 4 ≡ meff ṡ2
                                       2           2         2

   where the 1 + 1/2 represents the translational and rotational kinetic energy of the wheels, and the
   4 represents the kinetic energy of the board, since it travels at twice the speed as the centers of the
   wheels. On the other hand, the potential energy is

                                     V = −mgs sin θ(1 + 2) ≡ −Feff s

   where the 1 and 2 represent the potential energies of the wheels and board. Then we have
                                           Feff   3mg sin θ  6
                                    s̈ =        =           = g sin θ.
                                           meff   (11/2)m    11

   The acceleration of the board is twice this, giving the same answer,

                                             a = 2s̈ =      g sin θ.

      You might notice that as θ → π/2, the downward acceleration of the board becomes larger than g.
   This is the same phenomenon as in problem 8. Different parts of the system have different vertical
   accelerations, so it’s possible for part of it to fall faster than g. In this case, the extra downward

                   01W
   force on the board is due to friction from the wheels.


[User]
In the lab frame, a horizontal stick of proper length L has horizontal speed v. There
    is a horizontal thin sheet which has a hole of length L. Since the stick’s length is contracted to L/γ,
    it easily passes through the hole in the sheet, if the sheet is moved vertically. But in the frame of
    the stick, the sheet is moving horizontally, so the hole is length contracted instead. Qualitatively
    explain how the stick can still pass through the hole in this frame, in the following two cases:

      (a) The sheet has a uniform vertical velocity in the lab frame.

      (b) The sheet begins at rest at the lab frame, but is pushed upward a small amount when the
          stick passes over the hole, then ends at rest again.

Solution:
The idea behind this classic problem was first proposed by Rindler in 1961, then refined
    by Shaw in 1962, and incorporated into many textbooks. A detailed solution of Rindler’s original
    version, with illustrations, is given in section 6.4 of Understanding Relativity by Sartori.

      (a) The resolution is that the sheet is not horizontal in the stick’s frame. The simplest way to see
          this is to let the z-axis be vertical, and consider when different points in the sheet cross the
          point z = 0. In the lab frame, these events are all simultaneous, so they’re not simultaneous in
          the stick’s frame, which means that in the stick’s frame the sheet is rotated; you can calculate
          the angle with the Lorentz transformation. (This rotation is closely related to the Thomas
          precession effect mentioned in problem 13.) Since the sheet isn’t horizontal, the hole passes
          around the stick at an angle, so it fits. For a detailed quantitative solution, see this paper.

      (b) The resolution is that the sheet is not straight in the stick’s frame. Again, the simplest way to
          see this is to consider when different points in the sheet start to be raised. In the lab frame,
          these events are all simultaneous, so they’re not simultaneous in the stick’s frame. At any
          given moment in the stick’s frame, part of the sheet is still at the lower position, part of the
          sheet is already at the upper position, and part in between is moving upward while slanted,
          as in part (a). So, as the sheet moves horizontally, the hole appears to “bend around” the
          stick, letting it pass through.

         This might seem very disturbing. The sheet is always perfectly straight in the lab frame, but
         it has two kinks in the stick’s frame! But this is no more paradoxical than length contraction
         is. To decide whether a piece of an object is actually deformed, we need to look at it in its
         rest frame. An rod that’s severely length contracted in one frame is in no danger of breaking,
         and this sheet, which is severely kinked in some frames, is in no danger of tearing.
         Here’s another example: suppose you have a uniformly rotating cylinder. Then in the frame
         of an object moving along the axis of the cylinder, the cylinder is twisted because of the loss
         of simultaneity effect. (But it’s not really twisted, in the sense that if you work in the frame
         locally moving with any piece of the cylinder, it will have no shear stress.)
         The lesson is that the classical definition of a rigid body from M8, i.e. that angles and lengths
         between points on the body always remain the same, doesn’t work in special relativity; even
         if those conditions hold in one frame, they won’t necessarily in another.


[User]
Consider a particle of mass m confined to a one-dimensional box of length L. We consider the
quantum mechanics of this system. For simplicity, express your answers in terms of the quantity
α = h2 /8m as much as possible.

  1. In each energy level, the particle may be represented by a standing wave, where the wavefunc-
     tion is zero at the walls. Find the wavelength λ for the nth energy level.

  2. Using the de Broglie relation p = h/λ, find the energy En of the nth energy level.

  3. Electrons are fermions, meaning that each energy level can only be occupied with two electrons
     (one with spin up, and one with spin down). Let there be N electrons of mass m in this box,
     where N is an even number. Find the lowest possible total energy U0 of the system, i.e. the
     ground state energy of the system. You may neglect the Coulomb interaction between the
     electrons, and you may use the identity
                                      m
                                      X            m(m + 1)(2m + 1)
                                            n2 =

                                      n=1

  4. Write the total energy U1 in terms of U0 and relevant quantities when the system is in the
     first excited state. Do the same for the total energy U2 of the second excited state.

  5. In the ground state, find the magnitude of the force F on each wall in terms of U0 .

  6. In some astrophysical systems, the size L is free to vary, and in equilibrium the outward force
     found in part (5) balances the inward gravitational force. We can get a rough estimate for
     the equilibrium value of L by equating the total energy U0 to the total gravitational potential
     energy. Make a very rough estimate for the gravitational potential energy using dimensional
     analysis, treating the box as having uniform density. Equate this to U0 and find a rough
     estimate for the equilibrium length L in terms of N and other quantities.

Part B

Solution:
Consider a particle of mass m confined to a one-dimensional box of length L. We consider the
quantum mechanics of this system. For simplicity, express your answers in terms of the quantity
α = h2 /8m as much as possible.

  1. In each energy level, the particle may be represented by a standing wave, where the wavefunc-
     tion is zero at the walls. Find the wavelength λ for the nth energy level.

  2. Using the de Broglie relation p = h/λ, find the energy En of the nth energy level.

  3. Electrons are fermions, meaning that each energy level can only be occupied with two electrons
     (one with spin up, and one with spin down). Let there be N electrons of mass m in this box,
     where N is an even number. Find the lowest possible total energy U0 of the system, i.e. the
     ground state energy of the system. You may neglect the Coulomb interaction between the
     electrons, and you may use the identity
                                       m
                                       X            m(m + 1)(2m + 1)
                                             n2 =

                                       n=1

  4. Write the total energy U1 in terms of U0 and relevant quantities when the system is in the
     first excited state. Do the same for the total energy U2 of the second excited state.

  5. In the ground state, find the magnitude of the force F on each wall in terms of U0 .

  6. In some astrophysical systems, the size L is free to vary, and in equilibrium the outward force
     found in part (5) balances the inward gravitational force. We can get a rough estimate for
     the equilibrium value of L by equating the total energy U0 to the total gravitational potential
     energy. Make a very rough estimate for the gravitational potential energy using dimensional
     analysis, treating the box as having uniform density. Equate this to U0 and find a rough
     estimate for the equilibrium length L in terms of N and other quantities.

Solution. This is INPhO 2019, problem 2. The official solutions are here, an outline is:

  1. (2) λ = 2L/n for n ≥ 1.

  2. (3) The de Broglie relation p = h/λ means that

                                                     p2   n2 h2   αn2
                                        En =            =       =     .
                                                     2m   8mL2    L2

  3. (4) The energy levels from n = 1 up to n = N/2 are all occupied by two electrons. Thus,
                                            N/2
                                   2α X 2   α
                               U0 = 2  n =      N (N + 1)(N + 2).
                                   L       12L2
                                            n=1

  4. (6) To get to the first excited state, we remove one electron from the n = N/2 state, and put
     it in the n = N/2 + 1 state. That means
                                               2                      2
                                   α        N            α        N                    α
                       U1 = U0 −                     +              +1        = U0 +      (N + 1).
                                   L2       2            L2       2                    L2

     To get to the second excited state from the first excited state, we can then remove one electron
     from the n = N/2 − 1 state and put it in the n = N/2 state, since it now has an empty slot.
     The net change from the ground state is that one electron has been moved from n = N/2 − 1
     to n = N/2 + 1, giving
                                                    2                      2
                                   α        N                 α        N                    α
                       U2 = U0 −              −1          +              +1        = U0 +      2N.
                                   L2       2                 L2       2                    L2

  5. (2) By the definition of force,
                                                          dU0   2U0
                                                F =−          =     .
                                                          dL     L
  6. (3) Our extremely rough estimate is

                                                     GMtot   GN 2 m2
                                            UG ∼           ∼         .
                                                      L        L
     Setting this equal to U0 ∼ αN 3 /L2 , we have

                                                              αN
                                                         L∼       .
                                                              Gm2

Part B


[User]
\begin{wrapfigure}[5]{r}{0.35\linewidth}
	\vspace{-13pt}
	\includegraphics[width=\linewidth]{2017-lahg-07-res-cap2}
\end{wrapfigure}
The distance between the plates of a parallel plate capacitor is $d$. The capacitor’s capacity is $C$. Two dielectric plates of width $d/2$ are placed between the capacitor’s plates. The relative permittivity of one plate is $\varepsilon$, the other’s is $2\varepsilon$. What is the capacity of the capacitor now? What is the charge at the interface between the two plates if a voltage $U_0$ is applied to the capacitor?

Solution:
Vaatleme kahe dielektrikuga kondensaatorit kui kahte järjestikku ühendatud plaatkondensaatorit. Selleks asetame mõttelise metallplaadi dielektriliste kihtide eralduspinnale. Ühel kondensaatoril on seal laeng $+q$, teisel $-q$, seega vaba laeng on mõttelisel metallplaadil null (nii nagu vaja, sest tegelikult seal ju metallplaati pole). Algse dielektrikuta kondensaatori mahtuvus $C=\varepsilon_0S/d$. Mõtteliste kondensaatorite mahtuvused on $C_1=S\varepsilon_0\varepsilon/(d/2)=2\varepsilon C$ ja $C_2=S\varepsilon_02\varepsilon/(d/2)=4\varepsilon C$, seega summaarse mahtuvuse pöördväärtus
\[
C'=\frac{C_1C_2}{(C_1+C_2)}=\frac 43\varepsilon C.
\]

Dielektrikuga kondensaatori mahtuvus erineb dielektrikuta kondensaatori mahtuvusest, sest lisaks vabale laengule $q$ metallplaadil on veel dielektriku pinnal vastasmärgiline dielektriku polarisatsioonist tingitud (dielektrikust mitte-eraldatav) laeng $-q'$. Summaarne laeng $Q=q-q'$ on selline, nagu on antud pinge puhul samasugusel, ilma dielektrikuta kondensaatoril. Olgu ühe mõttelise kondensaatori kogulaeng (vaba laeng + dielektriku laeng) $Q_1=q-q_1'$ ja teisel $Q_2=q-q_2'$, siis sellel plaadil, mis jääb dielektrikute eralduspinnale, on kondensaatori plaatide laengumärke arvestades kogulaeng $Q_1-Q_2=q_2'-q_1'$. Seega, dielektrikute eralduspinnal oleva kogulaengu $q_2'-q_1'$ saame leida, kui selliste dielektrikuta kondensaatorite laengute vahe, mille pinge võrdub vastavale dielektrikukihile langeva pingega.

Mõttelistel kondensaatoritel on ühesugune laeng, seose $U=q/C$ tõttu on pinge pöördvõrdeline mahtuvusega, st ühele poole (kus dielektriline läbitavus on $\varepsilon$) langeb pinge $2U_0/3$ ja teisele poole (kus dielektriline läbitavus on $2\varepsilon$) $U_0/3$. Kui dielektriku kihti ei oleks, siis oleks kummagi mõttelise kondensaatori mahtuvus $2C$. Seega laeng mõttelistel kondensaatoritel oleks vastavalt $Q_1=4U_0C/3$ ja $Q_2=2U_0C/3$, mistõttu kogulaeng dielektrikute eralduspinnal
\[
q'=Q_1-Q_2=2CU_0/3.
\]
\vspace{0.5\baselineskip}

\emph{Alternatiivne lahendus}\\
Olgu kondensaatori plaadil laeng $q$, siis Gaussi teoreemist kondensaatori plaadi jaoks saame seose $SD=q$, millest $D=q/S$. Dielektrikute eralduspinnal on $D$-vektori normaalkomponent pidev, seega piirkonnas, kus dielektriline läbitavus on $\varepsilon$, on $E_1=D/\varepsilon\varepsilon_0$, teises piirkonnas $E_2=D/2\varepsilon\varepsilon_0$. Järelikult
\[
U_0=E_1\frac d2+E_2\frac d2=(Dd/2\varepsilon\varepsilon_0)(1+\frac 12)=\frac 34Qd/\varepsilon\varepsilon_0S,
\]
millest
\[
C'\equiv Q/U_0=\frac 43\varepsilon\varepsilon_0S/d=\frac 43C\varepsilon.
\]
Laengu piirpinnal $q'$ leiame Gaussi teoreemi abil mõttelise \enquote{karbi} jaoks, mis hõlmab piirpinda:
\[
q'=(E_1-E_2)\varepsilon_0S=DS/2\varepsilon=q/2\varepsilon=U_0C'/2\varepsilon=\frac 23CU_0.
\]

Kondensaator

\begin{wrapfigure}[5]{r}{0.35\linewidth}
	\vspace{-13pt}
	\includegraphics[width=\linewidth]{2017-lahg-07-res-cap2.pdf}
\end{wrapfigure}

Plaatkondensaator mahtuvusega $C$ omab plaatidevahelist kaugust $d$. Plaatide vahele asetatakse kaks dielektrilist plaati paksusega $d/2$. Ühe plaadi dielektriline läbitavus on $\varepsilon$, teisel $2\varepsilon$. Milline on nüüd kondensaatori mahtuvus? Milline laeng on plaatide eralduspinnal, kui kondensaatorile rakendatakse pinge $U_0$?

Mahtuvuse leidmiseks võib kondensaatorit vaadelda kui kahte jadamisi ühendatud dielektrikuga kondensaatorit. Selleks asetame mõttelise metallplaadi dielektrilise kihi eralduspinnale. Eralduspinna laengu leidmiseks võib rakendada Gaussi teoreemi mõttelise \enquote{karbi} jaoks, mis hõlmab piirpinda.


[User]
1. ANNIHILATION (6 points)

(by L. Franti)
An electron with kinetic energy 1MeV
travels along the z-axis and collides with a
positron at rest. The particles annihilate pro-
ducing a pair of two photons A and B with
equal energies.
i) (1.5 points) Find the speed of the electron
ve in the lab frame.
ii) (0.5 points) Find the energy Eγ of photon
A in the lab frame.
iii) (1 point) Find the momentum pγ of
photon A in the lab frame.
iv) (2 points) Find the angle α between the
z-axis and the momentum of photon A in the
lab frame.
v) (1 point) Why is it not possible for a
collision process to produce just one single
photon?
The rest mass of electron me =511keV/c2
.
You can express your results in electron volts
and related units.

Solution:
1. ANNIHILATION

i) Electron’s total energy E = γmec2
=⇒
γ = E
mec2 , where E = T + mec2
(here, T =
1MeV is the given kinetic energy). Now,
γ = 1/
q
1−v2
e/c2 =⇒ ve = c
p
1−1/γ2 =
c
q
1−m2
ec4/E2 =
c
p
T2+2Tmec2
T+mec2 . Numerically,
ve ≈0.941c.
ii) The photons fly away symmetrically
with respect to the electron’s trajectory: in
the zero-total-momentum frame, momentum
conservation implies that the photons have
equal momenta and, thus, equal energies,
and fly in exactly opposite directions; their
energies can be equal also in the positron’s
frame only if they are flying totally symmet-
rically. Each photon gets a half of the total
energy in the system: Eγ = 1
2(T +2mec2
) ≈
1.01MeV.
iii) Eγ = pγc =⇒ pγ = Eγ/c =1.01MeV/c.
iv) From E = p2
ec2
+ m2
ec4
, the elec-
tron’s momentum is pe =
q
E2
c2 −m2
ec2 =
1
c
p
T2 +2Tmec2. (Equivalent result can be
derived from pe = γmeve.) Momentum con-
servation in the z-direction then implies
that pe = 2pγ cosα =⇒ α = arccos
pe
2pγ
=
arccos 1 p
1+2mec2/T
=arctan
q
2mec2
T ≈45.3°.
v) In the center-of-mass frame, the total mo-
mentum of any system is zero. This means
that if the outcome of a collision is only a
single particle, then the particle’s momentum
must be zero in the center-of-mass frame.
However, a photon’s momentum can never
be zero, because otherwise it would have zero
energy and an infinite wavelength.

ANNIHILATION


[User]
Mechanical accelerator
A massless thread makes N turns around statically fixed cylinder, as shown in the figure. Initially,
the free (unwound) ends of the thread are parallel to the axis X. Then, a heavy point-like object
P is attached to one end of the thread while the other end is pulled with a constant velocity u
along X. Find the maximum velocity attained by the heavy object.

The thread is inextendable and flexible. Suppose that the turns of the thread are wound tightly
to one another and are placed practically in the same plane, perpendicular to the cylinder axis.
Neglect any friction in the system. Do not consider the force of gravity.

Solution:
T2: Mechanical accelerator

φ
Relative to Earth
lP
P
R
r
P

v
Rotating system of reference
Q
Figure 2: Mechanical accelerator
Solution I. Part 1: Thread in contact with the cylinder.
ThevelocityofthemassP canbedecomposedintolongi-
tudinal component vl along the thread, and a transverse
component v⊥ perpendicular to the thread:
⃗ v = vl⃗ e1 + v⊥⃗ e2
where the unit vectors ⃗ e1 and ⃗ e2 are parallel and per-
pendicular to the thread, respectively (see Fig. 2). Since
the thread is inextensible, the longitudinal component is
constant: vl = −u, i.e.
⃗ v = −u⃗ e1 + v⊥⃗ e2
The acceleration of P is, respectively:
⃗ a =
d⃗ v
dt
= −u
d⃗ e1
dt
+ v⊥
d⃗ e2
dt
+
dv⊥
dt
⃗ e2
EuPhO-2020 Online Theoretical Solutions. Language: English
Vectors ⃗ e1 and ⃗ e2 form a coordinate system, which ro-
tates as a rigid object with an angular velocity:
⃗ ω =
dφ
dt
⃗ e3
where ⃗ e3 = ⃗ e1 × ⃗ e2 is a unit vector perpendicular to the
plane of motion, i.e. along the cylinder axis, and φ is the
angle between the thread and the X-axis. Therefore, the
time derivatives of the basis vectors are:
d⃗ e1
dt
= ⃗ ω ×⃗ e1 =
dφ
dt
⃗ e2
and
d⃗ e2
dt
= ⃗ ω ×⃗ e2 = −
dφ
dt
⃗ e1
In this way, the acceleration of P can be represented in
terms of the angular velocity:
⃗ a = −v⊥
dφ
dt
⃗ e1 +

−u
dφ
dt
+
dv⊥
dt

⃗ e2
The only force, acting on P, is the tension of the thread.
Therefore, thecomponentoftheaccelerationperpendic-
ular to the tread, i.e. along ⃗ e2, is null:
−u
dφ
dt
+
dv⊥
dt
= 0
After integration over time, we obtain a relationship be-
tween the transverse velocity, acquired by P, and the an-
gle of rotation of the thread:
v⊥ = uφ
The end of the tread turns at a total angle of 2πN until
the tread detaches from the cylinder completely. There-
fore, the transversecomponent of the velocity of P at the
moment of detachment is:
v⊥ = 2πNu
and the magnitude of velocity:
v =
q
v2
l + v2
⊥ = u
p
(2πN)2 + 1
Part 2: Thread detached fom the cylinder. This ex-
pression, however, still does not represent the maxi-
mum velocity attained by P. In the frame of reference
of the free end of the thread, the mass continues to ro-
tate about the end of the thread. The velocity of P, rel-
ative to Earth, reaches maximum in the moment when
the thread reaches right angle with X-axis, i.e. the trans-
verse component of velocity of P aligns with ⃗ u:
vmax = u(2πN + 1)
Solution II. Part 1: Thread in contact with the cylin-
der. Consider a point Q on the end of the thread be-
ing pulled that coincided with P at the moment when
it touched the cylinder. Consider motion of the thread
in a system of reference (SR), which rotates at angular
speed ω = u/R around the center of the cylinder. In that
SR the part of the thread in contact with the cylinder is
at rest and the point Q rotates around the cylinder with
the angular velocity ωQ = −ω (see the figure).
Since the middle part of the thread is at rest, energy
of the mass P is conserved. For the same reason, the ve-
locity vP of P is perpendicular to the thread. Therefore
the kinetic energy acquired by the mass in the rotating
frame is equal to the decrease of its centrifugal potential
energy:
1
2
mv2
P = −
1
2
mω2
R2
+
1
2
mω2
r2
=
1
2
mω2
l2
P
where lP is the length of the unwound part of the thread
on the side of the mass P (see the figure). Therefore,
the mass P rotates around the fixture point of the thread
with a velocity:
vP =
u
R
lP
and a constant angular velocity:
ωP =
u
R
= ω.
Since ωP = −ωQ, in the rotating SR the two ends of the
thread will unwind symmetrically and the lengths of the
two straight parts of the string will be equal at any mo-
ment of time. Therefore, at the moment of detachment:
lP =
1
2
(2πRN) = πNR
and the detachment velocity of P is, respectively:
vP = πNu
When transforming the velocity of P to the Earth’s SR,
the velocity ⃗ vP should be added to the rotational veloc-
ity ⃗ ω × ⃗ r. It is easy to establish that the result for the
transverse component of P is:
v⊥ = 2vP = 2πNu
Part 2: Thread detached from the cylinder. In that part
we proceed exactly as in Part 2 of the first solution.
Solution III. Part 1: The thread in contact with the
cylinder
Like in Solution I we decompose the velocity into lon-
gitudinal and transverse components, and come to the
conclusion that the longitudinal component is vl =
−u. Afterwards, the acceleration of P is expressed. In
this case, however, we consider the longitudinal (cen-
tripetal) component of the acceleration:
al = −v2
⊥/l ≡ −v⊥
dφ
dt
From the second Newton’s law we obtain the tension F
of the thread:
F = −mv⊥
dφ
dt
The rate of change of the kinetic energy of the mass is
equal to the power of the tension force:
dEk
dt
= Fvl = +mv⊥
dφ
dt
u
Taking into account that:
Ek =
1
2
m(u2
+ v2
⊥)
EuPhO-2020 Online Theoretical Solutions. Language: English
and taking the first derivative from that expression, we
obtain: mv⊥dv⊥/dt = mv⊥dφ/dtu, or:
dv⊥
dt
= u
dφ
dt
From that point on we proceed exactly as in the first so-
lution.
Grading scheme: T2 part 1, Solution I
⃗ v is decomposed into vl and v⊥ 1.0 p
By condition of inextensibility vl = −u 1.0 p
Components of acceleration in terms of
dφ/dt
2.0 p
The tension force is along the thread and
a⊥ = 0
1.0 p
Finding dv⊥/dt = u · dφ/dt 1.0 p
At the moment of detachment v⊥ = 2πNu 1.0 p
Final v = u
p
1 + (2πN)2 1.0 p
Total for part 1: 8.0 p
Grading scheme: T2 part 1, Solution II
Introduction of rotational SR with angular
velocity ω = u/R
0.5 p
States that the wound string is at rest 0.5 p
Showing that the energy of P is conserved 1.0 p
Showing that the centrifugal force has po-
tential energy EP = −mω2
r2
/2
1.0 p
Conservation of energy equation 1.0 p
Proving that lP = lQ 0.5 p
Finding length lP = πNR at the moment of
release
0.5 p
Finding vP = πNu at the moment of re-
lease
1.0 p
Rotational velocity of the non-inertial SR
in point P ⃗ vrot = ⃗ ω ×⃗ r
0.5 p
Finding vrot⊥ = πNu 0.5 p
Final velocity relative to Earth v⊥ = 2πNu 1 p
Total for part 1: 8.0 p
Grading scheme: T2 part 1, Solution III
⃗ v is decomposed into vl and v⊥ 1.0 p
By condition of inextensibility vl = −u 1.0 p
Deriving v⊥ = ldφ/dt 0.5 p
Finding longitudinal acceleration al =
v2
⊥/l
0.5 p
Using Newton’s second law F = mal 0.5 p
Writing the kinetic energy Ek = m(u2
+
v2
⊥)/2
0.5 p
Using the work-energy theorem dEk/dt =
Fu
1.0 p
Deriving dv⊥/dt = udφ/dt 1.0 p
At the moment of detachment v⊥ = 2πNu 1.0 p
Final v = u
p
1 + (2πN)2 1.0 p
Total for part 1: 8.0 p
Grading scheme: T2 part 2, both solutions
Position in which maximum speed is
achieved
1.0 p
Value of vmax 1.0 p
Total for part 2 2.0 p

Mechanical accelerator


[User] Juku went sledging with his friends. When going back, two of Juku’s friends sat on his sledge and Juku tried to pull the sledge after him on a horizontal and snowy road. What is the minimal angle between the rope of the sledge and the ground so that it is possible for Juku to pull the sledge into motion? Juku’s mass is $m_1 = \SI{60}{kg}$ and the coefficient of friction between Juku’s boots and the snow is $\mu_1 = \SI{0.30}{}$. The mass of the sledge together with Juku’s friends is $m_2 = \SI{110}{kg}$ and the coefficient of friction between the sledge and the snow is $\mu_2 = \SI{0.20}{}$.

Solution:
Esiteks näeme, et kui Juku tõmbaks kelgunööri horisontaalselt, siis ei hakkaks kelk liikuma ükskõik kui suure tõmbejõu korral, sest Jukule mõjuv hõõrdejõud on väiksem kui kelgu liikumapanemiseks vajalik jõud:
\[
\mu_1 m_1 g < \mu_2 m_2 g.
\]
Esimesena hakkavad libisema hoopis Juku tallad.

Kui Juku tõmbab nööri teatud nurga all ülespoole, siis tekib nööris vertikaalne jõu komponent $F_{\mathrm{v}}$, mis tõstab kelku ülespoole ja surub Jukut allapoole. Seega mõjub Jukule tema libisemise piiril hõõrdejõud $F_{\mathrm{h}1} = \mu_1 \left(m_1 g + F_{\mathrm{v}}\right)$ ja kelgule tema libisemise piiril hõõrdejõud $F_{\mathrm{h}2} = \mu_2 \left(m_2 g - F_{\mathrm{v}}\right)$. Kuna küsiti minimaalset nurka, siis peab kelk olema libisemise piiri napilt ületanud ja Juku sellele napilt alla jääma ehk piirjuhul $F_{\mathrm{h}1} = F_{\mathrm{h}2} = F_{\mathrm{h}}$, kus $F_{\mathrm{h}}$ tähistab nööris tekkiva jõu horisontaalset komponenti.
Jõudude tasakaalu võrrandist
\begin{equation*}
\mu_1 \left(m_1 g + F_{\mathrm{v}}\right) = \mu_2 \left(m_2 g - F_{\mathrm{v}}\right)
\end{equation*}
saame avaldada nööris tekkiva jõu vertikaalse komponendi:
\begin{equation*}
\mu_1 m_1 g + \mu_1 F_{\mathrm{v}} = \mu_2 m_2 g - \mu_2 F_{\mathrm{v}},
\end{equation*}
\begin{equation*}
F_{\mathrm{v}} \left(\mu_1 + \mu_2\right) = g\left(\mu_2 m_2 - \mu_1 m_1\right),
\end{equation*}
\begin{equation*}
F_{\mathrm{v}} = g\frac{\mu_2 m_2 - \mu_1 m_1}{\mu_1 + \mu_2}.
\end{equation*}

Jõu horisontaalkomponendi leidmiseks asendame saadud tulemuse näiteks Jukule mõjuva hõõrdejõu võrrandisse
\begin{equation*}
F_{\mathrm{h}} = \mu_1 \left(m_1 g + g\frac{\mu_2 m_2 - \mu_1 m_1}{\mu_1 + \mu_2}\right)
\end{equation*}
ja avaldame:
\begin{align*}
F_{\mathrm{h}} &= g \mu_1 \frac{m_1 \left(\mu_1 + \mu_2\right) + \mu_2 m_2 - \mu_1 m_1}{\mu_1 + \mu_2} = \\
&= g \mu_1 \frac{\mu_2 m_1 + \mu_2 m_2}{\mu_1 + \mu_2} = g\mu_1\mu_2\frac{m_1+m_2}{\mu_1+\mu_2}.
\end{align*}

Nurk kelgunööri ja maapinna vahel on
\begin{equation*}
\alpha = \arctan\left(\frac{F_{\mathrm{v}}}{F_{\mathrm{h}}}\right) = \arctan\left(\frac{\mu_2 m_2 - \mu_1 m_1}{\mu_1 \mu_2 \left(m_1 + m_2\right)}\right) = \ang{21}.
\end{equation*}

Kelk

Juku läks sõpradega kelgutama. Teel tagasi istusid Juku kaks sõpra kelgule ja Juku üritas kelku horisontaalsel lumisel teel enda järel vedada. Kui suur on minimaalne kelgunööri nurk maapinnaga, mille korral on Jukul võimalik kelk liikuma tõmmata? Juku mass $m_1 = \SI{60}{kg}$ ja hõõrdetegur Juku saabaste ning lume vahel $\mu_1 = \SI{0.30}{}$. Kelgu mass koos Juku sõpradega $m_2 = \SI{110}{kg}$ ja hõõrdetegur kelgu ning lume vahel $\mu_2 = \SI{0.20}{}$.

Mõistlik on nööri pinge jaotada horisontaalseks ja vertikaalseks komponendiks $T_x$ ja $T_y$. Sellisel juhul kasvab Juku saabastele mõjuv toereaktsioon $T_y$ võrra ning kelgule mõjuv toereaktsioon kahaneb $T_y$ võrra. Minimaalse kelgunööri nurga puhul on Jukust tulenev hõõrdejõud võrdne kelgu hõõrdejõuga.


[User]
Let’s check the consequences of this idea.

      (a) We should be able to take the partial derivatives of U with respect to S or V in either order.
          By combining this idea with your solution to problem 1, infer a nontrivial relationship between
          the derivatives of T and P , and check that it is satisfied for an ideal gas.

     (b) If heat was a state function, we could take the partial derivatives of Q with respect to T or V
         in either order. Check that this implies a relationship which is not satisfied for an ideal gas.

Solution:
(a) Since we can take the partial derivatives of U with respect to S and V in either
         order, we conclude that we need the nontrivial relation

                                                 ∂T      ∂P
                                                      =−      .
                                                 ∂V S    ∂S V

         To evaluate the left-hand side, we need to consider how T varies with V in an adiabatic
         expansion. In this case T V γ−1 is constant, so that
                                                  dT           dV
                                                     = (1 − γ)
                                                  T            V
         from which we conclude that
                                                ∂T      T
                                                     = − (γ − 1).
                                                ∂V S    V
         To evaluate the right-hand side, note that for constant volume heating, dU = d̄Q = T dS, so
                                                                   −1
                                      ∂P         ∂P           ∂U
                                            =T          =T              .
                                      ∂S V       ∂U V         ∂P V

         But for an ideal gas, we have U = nRT /(γ − 1) = P V /(γ − 1), so

                                                 ∂P    T
                                                      = (γ − 1)
                                                 ∂S V  V

         which confirms the expected result.

     (b) By definition, the differential of heat is

                                                             ∂Q
                                             d̄Q = CV dT +        dV
                                                             ∂V T

         and the required identity, if heat were a state function, is

                                            ∂CV        ∂      ∂Q
                                                   =                  .
                                            ∂V T      ∂T V ∂V T

         The left-hand side just vanishes for an ideal gas. As for the right hand side, we note that

                                     ∂Q     ∂U     ∂W      ∂W
                                          =      −      =−      =P
                                     ∂V T   ∂V T   ∂V T    ∂V T

          where we used the fact that the internal energy of an ideal gas depends only on temperature.
          Thus, the right-hand side is
                                                ∂P       nR
                                                      =      ̸= 0
                                                ∂T V      V
          so the identity is not satisfied, as expected.


[User]
01^
                        USAPhO 2019, problem A2. An elegant, tricky radiation problem. (This is
    essentially the same problem as Physics Cup 2022, problem 5.)

Solution:
01^
                        USAPhO 2019, problem A2. An elegant, tricky radiation problem. (This is
    essentially the same problem as Physics Cup 2022, problem 5.)


[User] USAPhO 2016, problem A3. Print out the custom answer sheet before starting.

Solution: 01W   USAPhO 2016, problem A3. Print out the custom answer sheet before starting.


[User]
A uniform ring of mass m and radius R has a point mass of mass M attached to it. The ring is
placed on the ground, with the point mass initially at its highest point, and is given an infinitesimal
sideways impulse in the plane of the ring. Assume the ring does not slip.
   1. Assuming the ring never loses contact with the ground, find the angular velocity ω of the ring
      as a function of the angle θ through which it has rotated.
   2. Continuing to assume the ring never loses contact with the ground, find the vertical component
      of the force on the ring-mass system as a function of θ.
   3. It turns out that for some range of values of m/M , the ring leaves contact with the ground
      at some point. What are these values? (For partial credit, you can instead prove that there
      exists a value of m/M so that the ring jumps.)

Solution:
A uniform ring of mass m and radius R has a point mass of mass M attached to it. The ring is
placed on the ground, with the point mass initially at its highest point, and is given an infinitesimal
sideways impulse in the plane of the ring. Assume the ring does not slip.

   1. Assuming the ring never loses contact with the ground, find the angular velocity ω of the ring
      as a function of the angle θ through which it has rotated.

   2. Continuing to assume the ring never loses contact with the ground, find the vertical component
      of the force on the ring-mass system as a function of θ.

   3. It turns out that for some range of values of m/M , the ring leaves contact with the ground
      at some point. What are these values? (For partial credit, you can instead prove that there
      exists a value of m/M so that the ring jumps.)

Solution. This is a classic problem, most recently popularized by Tadashi Tokieda’s paper The
Hopping Hoop and originally published in John Littlewood’s Miscellany.

   1. (5) By a conservation of energy argument, and letting µ = m/M for simplicity,
                                            s
                                                 1 − cos θ g
                                         ω=                    .
                                               1 + cos θ + µ R

  2. (10) By differentiating the angular velocity, we have

                                                (1 + µ/2) sin θ g
                                         α=                       .
                                               (1 + cos θ + µ)2 R

    Deriving this is a little messy, and maybe the easiest way to do it is to note that d(ω 2 )/dt = 2ωα,
    since ω 2 doesn’t have a square root.
    Only the mass itself accelerates in the y direction, so the vertical force is M ay where ay is its
    acceleration. Using polar coordinates, the mass has

                                             y = R(1 + cos θ)

    which implies that
                             vy = −Rω sin θ,     ay = −R(α sin θ + ω 2 cos θ).
    Plugging in our earlier results, we find the messy result

                                         sin(θ/2)2
                 Fy = M ay = −M g                      (3 + µ + (4 + 3µ) cos θ + cos 2θ).
                                      (1 + µ + cos θ)2

  3. (10) The ring will leave contact with the ground if the expression above implies the normal
     force is negative, Fy < −M (1 + µ)g, which is equivalent to

                          sin(θ/2)2
                                        (3 + µ + (4 + 3µ) cos θ + cos 2θ) ≥ 1 + µ.
                       (1 + µ + cos θ)2
    It’s a little easier to understand this by working entirely in terms of cos θ. Using half-angle
    and double-angle identities gives
                  1 − cos θ        1
                                             (2 + µ + (4 + 3µ) cos θ + 2 cos2 θ) ≥ 1 + µ.
                      2     (1 + µ + cos θ)2

    The ring will indeed jump for small enough m/M . To see this, set m/M = 0 to get
                             1 − cos θ      1
                                                    (2 + 4 cos θ + 2 cos2 θ) ≥ 1
                                 2     (1 + cos θ)2
    which greatly simplifies to
                                               1 − cos θ ≥ 1.
    This condition is violated when θ > π/2, upon which the ring jumps off the ground.
    Quantitatively, the ring jumps when m/M < 1/13. By clearing denominators and using a
    series of trigonometric identities, we can show that jumping occurs at the moment that

                                                      µ(2 + µ)2
                                   µ + 2 cos θ +                    = 0.
                                                   (1 + µ + cos θ)2
    So to see if jumping does or does not occur, we need to minimize this with respect to θ.
    Jumping occurs if the minimum is negative, because then the expression would have crossed
    zero at some point. Setting the derivative with respect to cos θ equal to zero gives

                                      µ(2 + µ)2 = (1 + µ + cos θ)3 .

     Plugging this result back in gives the condition for jumping,

                           µ + 2((µ(2 + µ)2 )1/3 − µ − 1) + (µ(2 + µ)2 )1/3 ≤ 0.

     The threshold in µ for jumping occurs when this is equal to zero,

                                        2 + µ = 3µ1/3 (2 + µ)2/3 .

     Cubing both sides gives a cubic equation,

                                       13µ3 + 51µ2 + 48µ − 4 = 0.

     The solution we want is µ = 1/13, with the other two extraneous. One way you can arrive
     at this result is to notice that the cubic is negative at µ = 0 and positive at µ = 1/10, so it
     must cross zero somewhere in between. Then 1/13 is one of the only options permitted by
     the rational root theorem, and plugging it in and checking works.
     Of course, it’s crazy that all this is worth just 10 points! In practice, some points are indeed
     much harder to get than others. But the true purpose of this question is to test how you fare
     under extreme time pressure. If you spent a lot of time working on this part, but lost some
     much easier points on B1 or B2, you should reconsider your strategy for future mock exams.

    Tough question, right? But it’s actually even tougher than that: it turns out that for a massless
ring, a jump doesn’t happen at all! Instead, the ring “glides” on the ground, always touching it
but exerting zero normal force. For a massive ring, a jump can happen, but it must be preceded by
slipping, i.e. the analysis above doesn’t work. In fact, it took numerous papers arguing back and
forth to figure this out; the latest work (published just in 2019) is here.


[User]
Mööda liigendi abil seina külge kinnitatud väga pikka ja tühiselt
kerget varrast saab libiseda väike rõngas massiga $m$. Esialgu asub rõngas liigendist kaugusel $l$ ja varras on horisontaalne. Ajahetkel $t = \num{0}$ hakkab süsteem
vabalt liikuma. Leidke varda ja horisontaali vahelise nurga $\alpha$ ajaline sõltuvus.
Kõik liikumised lugeda hõõrdevabaks.

Solution:
\begin{center}
	\includegraphics[width = 0.5\linewidth]{2011-v3g-02-varras_lah.pdf}
\end{center}

Kuna varras on kaalutu, peab sellele mõjuv summaarne jõud olema \num{0}. Vastasel korral mõjuks sellele Newtoni III seaduse kohaselt lõpmatu jõud, mis pole füüsikaline. Sellest saab järeldada, et massile mõjuv normaaljõud on null ning hõõrdeta libisemise tõttu on ka vardaga paralleelne jõukomponent null. Seega ei mõju massile varda poolt ükski jõud ning mass on vabalanguses.

Aja $t$ jooksul jõuab mass langeda vahemaa $\frac{gt^2}{2}$ ning varda ja horisontaali vaheline nurk avaldub kui (vt joonist) $\tan\alpha = \frac{gt^2}{2l}$.

Varras

Mööda liigendi abil seina külge kinnitatud väga pikka ja tühiselt
kerget varrast saab libiseda väike rõngas massiga $m$. Esialgu asub rõngas liigendist kaugusel $l$ ja varras on horisontaalne. Ajahetkel $t = \num{0}$ hakkab süsteem
vabalt liikuma. Leidke varda ja horisontaali vahelise nurga $\alpha$ ajaline sõltuvus.
Kõik liikumised lugeda hõõrdevabaks.

Ülesanne näeb keerulisem välja, kui see tegelikult on. Olukorda lihtsustab oluliselt asjaolu, et varras on tühiselt kerge ja libisemised hõõrdevabad.


[User]
USAPhO 2011 A4 is a neat problem, but it has some typos and unphysical assump-
    tions; this is an improved version. Consider a simplified model of the electromagnetic radiation
    inside a cubical metal box of side length L at temperature T . In this model, modes of the electric
    field have spatial dependence

                                  E(x, y, z) = E0 sin(kx x) sin(ky y) sin(kz z)

    where one corner of the box lies at the origin and the box is aligned with the x, y, and z axes. For
    simplicity, we will treat the electric field as a scalar, even though it actually has direction.

      (a) The electric field must be zero everywhere at the sides of the box. What condition does this
          impose on the ki ?

     (b) Show that for a typical box at room temperature, L ∼ 1 m and T ∼ 300 K, the quantity kB T
         is much greater than ℏω0 , where ω0 is the lowest nonzero mode angular frequency. Therefore,
         for the rest of the problem, you may approximate kB T ≫ ℏω0 .

      (c) Each permitted value of the triple (kx , ky , kz ) corresponds to a mode, which can be occupied
          by any number of photons. Each photon has an energy E = ℏω, where ω = ck is the angular
          frequency of the mode. How many modes have an energy per photon of at most kB T ?

     (d) As a crude approximation, suppose that in thermal equilibrium, each mode with energy per
         photon at most kB T contains exactly one photon, while all other modes contain no photons.
         Compute the total energy of the photons in the box. (Answer: (kB T )4 L3 /8π 2 ℏ3 c3 .)

    Note that the procedure
                   H         here is different from what we did above. Before, we started with particles
    and quantized p dx to get the allowed quantum states. Here, we’re treating a situation with many
    particles (photons), which are excitations of an underlying field (the electromagnetic field). In this
    case, we found the (normal) modes of the classical field, then quantized by saying that photons
    could occupy these modes. This is the methodology of quantum field theory.

Solution:
(a) We see that kx L = nx π, ky L = ny π, and kz L = nz π.

 (b) Plugging in the numbers, we have kB T /ℏω0 ∼ 4 × 104 ≫ 1, as desired.

 (c) Define the vector n = (nx , ny , nz ). For each mode, the energy per photon is

                                                            πℏc q 2
                                       E = ℏω = ℏck =            nx + n2y + n2z .
                                                             L
     Moreover, the values of the ni are positive integers. Therefore, the quantum states we are
     looking for occupy an eighth of a sphere in phase space, bounded by states with
                                                                Emax L
                                                nmax =                 .
                                                                 πℏc
     Because kB T ≫ ℏω0 , there are many such modes, so we can find the answer approximately
     by computing the volume of this region. The number of modes (i.e. the number of values of
     n) with an energy per photon of at most Emax is thus
                                                                                3
                                               1       4 3                 1         EL
                                   N (Emax ) =          πn               = 2                   .
                                               8       3 max              6π         ℏc

     In this case, the requested answer is
                                                                              3
                                                       1              kB T L
                                           N (kB T ) = 2                            .
                                                      6π               ℏc

     Note that this is different from the official USAPhO solution, because they also allowed
     negative values for the ni . This is incorrect, because flipping the sign of one of the ni gives
     you exactly the same mode, up to an irrelevant −1 factor.

 (d) We sum over the modes. For each occupied mode, we assume the energy stored is E, so
                              Z kB T                                 3 Z kB T
                                                                                1 (kB T )4 L3
                 Z
                                         dN      1              L
            U=       E dN =            E    dE = 2                               E(3E 2 ) dE =                    .
                               0         dE     6π              ℏc        0                        8π 2 ℏ3 c3


[User]
Consider a spherical cap that is formed by slicing a sphere of radius R by a plane,
    so that the altitude from the vertex to the base is h. Find the area of its curved surface using an
    appropriate integral.

    7      Approximating Integrals

Solution:
This is a double integral, where it’s best to use spherical coordinates. Recall that the
    volume element in spherical coordinates was dV = r2 dr sin ϕ dϕ dθ. Thus, the area element for a
    part of this sphere is dA = R2 sin ϕ dϕ dθ. The area integral is
                                      Z cos−1 ((R−h)/R)          Z 2π

                              A=R                       sin ϕ dϕ      dθ = 2πhR.
                                       0                               0
    After doing the trivial inner integral, this approach is just slicing the surface by dϕ. You can also
    equivalently solve it by slicing it in dz. In that case the integrand is a bit more complicated, but
    the bounds are simpler.

    7    Approximating Integrals


[User]
Superconducting mesh
Consider a mesh made from a flat superconducting sheet by drilling a dense grid of small holes into
it. Initially the sheet is in a non-superconducting state, and a magnetic dipole of dipole moment
m is at a distance a from the mesh pointing perpendicularly towards the mesh. Now the mesh is
cooled so that it becomes superconducting. Next, the dipole is displaced perpendicularly to the
surface of the mesh so that its new distance from the mesh is b. Find the force between the mesh
and the dipole. The pitch of the grid of holes is much smaller than both a and b, and the linear
size of the sheet is much larger than both a and b.

Solution:
Solution - Superconducting mesh
    The most important physics to consider is that the magnetic flux through the superconducting
mesh is effectively locally locked in place. Consider this effect before anything else. Once the mesh
is cooled to the superconducting state the magnetic field as a function of position on the mesh
cannot be varied, regardless of the change in location of the dipole. Since the magnetic field is
effectively specified along this superconducting plane, the problem reduces to a boundary value
problem that is traditionally solved by the method of images.

    First, consider what happens if the physical dipole is moved far away from the mesh. An image
dipole must be located that fixes the magnetic field to be unchanged. This can be done with an
image dipole that is located a distance a behind the mesh, and it must have the same orientation
m. Now bring back the original dipole, placing it a distance b. It is necessary to cancel out the
field from this original, but now displaced, dipole with an opposite dipole −m placed behind the
mesh at a distance b.

    Double check your work. If the original dipole is placed at the original location a, then there
is no need for image charges, and they should cancel out. Indeed, the two image dipoles will, as
they have opposite orientations.

   The force between the dipole and the image charges must be determined. Though it might be
possible to write down these answers quickly, the derivation is shown below.

  Consider first that a magnetic dipole moment m can be thought of as a pair of magnetic
monopoles of strength qm and −qm separated by a distance d such that m = qm d. Determine the
magnetic field strength a distance x fl d away from the dipole:

                                          μ0 qm μ0 (−qm )
                                     B=         +
                                          4π x2   4π (x + d)2

    It should be clear that qm is at the origin and −qm is a distance d farther away from the
reference point x where the field B is being determined. This expression is exact.
    The second term can be subjected to a binomial expansion and then:

                              μ0 qm    μ0 qm       d      μ0 qm d   μ0 m
                        B≈        2
                                    −       2
                                               1−2     =       3
                                                                  =
                             4π x      4π x        x     2π x       2π x3

    Now consider the magnetic force on a dipole at the location x in a non-uniform field B, which
is given by:

                                   F = −qm B(x) + qm B(x + d)

which can be approximated by a Taylor expansion of B,

                                                          dB
                           F ≈ −qm B(x) + qm B(x) + d
                                                          dx x
                                                     3μ0 m2

                                        3μ0 m
                           F = qm d −           =  −
                                        2π x4        2π x4

The negative sign means that two parallel identical dipoles separated by a distance x will attract.
   Returning to the problem, the physical dipole at b will be attracted to the image dipole at
location −a and repelled from the image dipole at −b, so:

                    3μ0 m2         3μ0 m2          3μ0 m2

                                                              1        1
                  −              +              =                −            ,
                     2π (b + a)4    2π (b + b)4      2π      16b4 (a + b)4

where a negative sign means that the physical dipole feels attraction toward the mesh.
   It is entertaining to consider what happens if b is almost the same as a, say b = a + δ. In this
case,

                                   3μ0 m2

                                                1            1
                              F =                      −             ,
                                     2π     16(a + δ)4 (2a + δ)4

or

                              3μ0 m2 1

                                                    δ             δ/2
                          F ≈                   1−4         − 1−4        ,
                                2π 16a4             a              a

which simplifies further into:

                                                  3μ0 m2
                                          F ≈−           δ
                                                  16πa5

Now to interpret. A negative force here is a force of attraction toward the mesh. A positive δ
is moving the physics dipole away from the mesh. As such, the force is a linear restoring force,
and slight disturbances to the physical dipole will result in simple harmonic oscillations about the
original position.

Grading Scheme
 1.5 Recognition of nature of problem

        − Recognize flux trapping in superconductor (1.0 pt)
        − Recognize that flux trapping creates a boundary value problem (0.5 pt)

 4.0 Recognize that the boundary value problem requires two image dipoles

        − First image dipole to create original B field on mesh (0.5 pt)
        − Correct location of first image dipole (0.5 pt)
        − Correct magnitude of first image dipole (0.5 pt)
        − Correct orientation of first image dipole (0.5 pt)
        − Second image dipole to cancel new B field on mesh (0.5 pt)
        − Correct location of second image dipole (0.5 pt)
        − Correct magnitude of first image dipole (0.5 pt)

       − Correct orientation of first image dipole (0.5 pt)

 2.0 Determine the force between two dipoles

       − Determine B field a distance from a dipole (1 pt)
       − Determine force on a dipole in a non-uniform B field (1 pt)

 2.5 Determine the force between the physical dipole and the mesh

       − Correct magnitude and direction of force from image dipole one (1 pt)
       − Correct magnitude and direction of force from image dipole two (1 pt)
       − Correct magnitude and direction of force (0.5 pt)

Some notes:

   • Dimensionally correct expression with no shown work but have wrong prefactor get zero
     marks

   • Dimensionally correct expression that show work but have wrong prefactor caused from clear
     trivial math mistake get 1/2 marks

   • Dimensionally correct expression that whow work but have wrong prefactor caused from
     serious math mistake or any physics mistake get zero marks

   • Dimensionally incorrect expression with no shown work get zero marks

   • Dimensionally incorrect answers that show work get zero marks

   • Follow on errors that use dimensionally correct, but wrong, derived inputs are not penalized
     further

   • Follow on errors that use dimensionally incorrect derived inputs are penalized half marks
     each time the input is used

   • Writing a formula incorrectly without showing the derivation and using it as an input is not
     considered a follow on error, but is instead a non-trivial error with a result of zero marks for
     that part

   • Ambiguous locations, magnitudes, or orientations receive zero marks

   • Correctly identifying the locations, magnitudes, or orientations of both of the two image
     dipoles without clearly specifying why will receive full 1.5 points for recognizing the nature
     of the problem

   • Correctly identifying the locations, magnitudes, or orientations of only one of the image
     dipoles without clearly specifying why will receive 0.5 points for recognizing the nature of
     the problem

Superconducting mesh


[User]
A uniform rope of mass m and length L hangs from a ceiling.

      (a) Show that the time it takes for
                                        pa transverse wave pulse to travel from the bottom of the rope
          to the top is approximately 2 L/g. Under what circumstances is this approximation good?

      (b) Does the pulse get longer or shorter as it travels?

Solution:
(a) The velocity is                p      p
                                               v=    T /µ = T L/m
          and T = xmg/L where x is the distance from the bottom, so
                                                           √
                                                      v=       xg.

          At the most naive level, a wave pulse just travels along the string with a varying velocity, and
          takes a total time                                                  s
                                     Z L          Z L            Z L
                                         dt            dx      1      dx        L
                                t=          dx =          =√         √ =2          .
                                      0 dx         0    v       g 0     x        g
          This approximation makes sense as long as the wave pulse can’t “see” that the velocity is
          actually changing, which means it works if the pulse has a length much smaller than the length
          of the rope itself. This idea is discussed further for the case of quantum matter waves in X1.

      (b) Because the tension is higher at higher points on the rope, the top part of the pulse is always
          traveling a bit faster than the bottom. Thus, the pulse gets longer as it travels up.


[User]
As a warmup, take the Conceptual Survey of Electricity and Magnetism test. This
    should take no longer than 1 hour. You do not have to justify your answers; just list them.

Solution:
As a warmup, take the Conceptual Survey of Electricity and Magnetism test. This
    should take no longer than 1 hour. You do not have to justify your answers; just list them.

    Answer. BABBC EBBBC EDEDA EEDAD EDACD AECCA ED


[User]
8. ELECTRIC FIELD (6 points)

(by
L. Franti) A conducting ring with radius R
carries a large current I. The ring lies sta-
tionary in the xy-plane with its centerpoint
at (0,0,0). An observer approaches the ring
parallel to the z-axis with velocity v and
measures the electric field at distance r from
the axis. Assume that v¿ c and r ¿ R.
i) (2 points) Find the direction and mag-
nitude of the magnetic field B(z) on z axis in
stationary reference frame.
ii) (4 points) Estimate the magnitude and
direction of electric field E(z,r) measured by
the observer when he is at a distance z from
the ring’s plane.

Solution:
8. ELECTRIC FIELD

i) By the Biot-Savart law, every piece of
the ring with length and direction d~ l cre-
ates a magnetic field d~ B =
μ0Id~ l×â
4πa2 , where ~ a
is the displacement vector from the piece to
the point where we are calculating ~ B, and
â = ~ a/a. By symmetry, the magnetic field
on the z-axis adds up to being exactly along
the z-axis, because the perpendicular com-
ponents cancel out. dBz is proportional to dl
and d~ l ⊥ â. The vector d~ l×â is at an angle
α = arctan r
z from the horizontal. Therefore
we can add up the contributions and write
B(z) =
μ0I×2πRsinα
4π(z2+R2)
=
μ0IR2
2(z2+R2)3/2 . The field is
along the z-axis.
ii) The electric field measured by a charged
observer must be in the same direction as
the Lorentz force acting on the observer in
the laboratory frame of reference. Therefore
the electric fieldlines are circles around the z-
axis. Now we may apply Faraday’s induction
law to such a circle, as it approaches the ring:
the total electromotive force E(z,r)×2πr =
Φ̇ = −dΦ
dz v ≈ − d
dz B(z)×πr2
×v =⇒ E(z,r) =
−rv
2 B0
(z)=
3μ0IrR2vz
4(z2+R2)5/2 .

ELECTRIC FIELD