[User]
Find the readings of the ammeter and the voltmeter in the circuit diagram given in the drawing. All the resistors have a resistance $R=\SI{1}{\ohm}$, voltage on the battery’s leads is $U=\SI{7}{\volt}$.
\begin{center}
	\includegraphics[width=0.4\textwidth]{2017-v3g-06-viisnurk}
\end{center}

Solution:
Joonistame lihtsustatud skeemi, arvestades et ampermeeter ja voltmeeter
on ideaalsed. See tähendab, et ampermeetri võib asendada juhtmega ja
voltmeetri lihtsalt kõrvaldada. Tähistame uue skeemi sõlmpunktid
tähtedega P --- patarei plussklemm; M --- patarei miinusklemm; A ---
ampermeetri asukoht; V --- voltmeetri alumine klemm. Takistid tähistame
numbritega ühest viieni, liikudes päripäeva ja alustades viisnurga
ülemisest tipust paremale jäävast takistist. Tulemus on toodud joonisel
oleval skeemil.
\begin{center}
	\includegraphics[width=0.5\textwidth]{2017-v3g-06-viisnurk-lah}
\end{center}
Kirchoffi vooluseaduse tõttu on ampermeetri vool võrdne
takistite 4 ja 5 voolude vahega, sest see osa voolust, mis läks läbi takisti 4, aga ei läinud läbi takisti 5, pidi järelikult läbi ampermeetri minema. Voltmeetri näit on võrdne klemmide
M ja V vahelise pingega.
Skeemis on takistid 1 ja 5 paralleelselt, seega on nende takistus kokku $\frac{1}{2}\si{\ohm}$. Takisti 4 on paralleelselt
takistite 2 ja 3 jadaühendusega, seega on nende takistus
\[
\frac{1}{\frac{1}{1\si{\ohm}}+\frac{1}{2\si{\ohm}}} = \frac 23 \si{\ohm}.
\]
Terve skeemi kogutakistus on seega
\[
R_K=\left(\frac 12 +
\frac 23\right)\si{\ohm}=\frac 76\si{\ohm},
\]
koguvool seega $I=\mathcal
E/R_K=\SI{6}A$. Takistitest 1 ja 5 läheb sama palju voolu läbi,
mistõttu takisti 5 vool on $I_5=\frac 12 I=\SI{3}A$ ja pingelang sellel takistil on \SI{3}{\volt}. Seega on punktide A ja P vahel pinge \SI{4}{\volt} ja vool läbi takisti 4
on $I_4=\frac{\SI{4}{\volt}}{\SI{1}{\ohm}}=\SI{4}A$. Ampermeetri
näit on seega $I_4-I_5=\SI 1A$.

Pinge M ja V vahel on patarei pingest väiksem takisti 3 pinge võrra. Kuna punktide A ja P vahel oli pinge \SI{4}{\volt}, siis arvestades, et takistid 2 ja 3 on samasugused ning jadamisi ühendatud, on takistil 3 pinge \SI{2}V.
See annab voltmeetri pingeks $\SI{7}V -\SI{2}V= \SI{5}V$.

Viisnurk

Leidke joonisel toodud skeemis ampermeetri ja voltmeetri näidud. Kõik takistid on takistusega $R=\SI{1}{\ohm}$, pinge patarei klemmidel $U=\SI{7}{\volt}$.

\begin{center}
	\includegraphics[width=0.4\textwidth]{2017-v3g-06-viisnurk}
\end{center}

Elektriskeemi käitumisest arusaamiseks tasub skeem selgemalt ümber joonistada, nii et ampermeeter on asendatud juhtmega ja voltmeeter lihtsalt kõrvaldatud.


[User]
In this problem we’ll apply the above results to balls of charge.
      (a) Compute the potential energy of a uniformly charged ball of total charge Q and radius R.

      (b) Show that the potential energy of two point charges of charge Q/2 separated by radius R is
          lower than the result of part (a).

      (c) Hence it appears that it is energetically favorable to compress a ball of charge into two point
          charges. Is this correct?

Solution:
(a) We can find the potential by building up the ball by placing charges from infinity.
         Consider a shell of charge at radius r, and let the charge density by ρ = Q/( 43 πR3 ). The
         energy needed to put the shell there is dU = kQenc dQ/r, where Qenc = 43 ρπr3 is the charge
         inside and dQ = 4ρπr2 dr is the charge in the shell added to the sphere. Then the energy
         needed to build the ball, which is the potential energy of the ball, is

                                                        3kQ2 R 4
                          Z R
                                  r3                                       3kQ2     3Q2
                                                              Z
                    Ua =      kQ 3 (3Qr2 dr/R3 )/r =              r  d r =       =         .
                           0      R                       R6 0              5R     20πϵ0 R

      (b) From U = kq1 q2 /r, we find that for two point charges the potential energy is

                                                       kQ2     Q2
                                               Ub =        =
                                                       4R    16πϵ0 R
          which is less than Ua .

      (c) It’s wrong because in part (b), the energy needed to create the point charges, by squeezing
          the two halves of the ball down, is not included. Plugging in a radius of zero into part (a), we
          see that this energy is actually infinite. (Of course, in reality it doesn’t take infinite energy
          to produce electrons, which are point charges. Classical electrodynamics breaks down when
          describing such a process, which can only be properly understood within relativistic quantum
          field theory.)

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


[User]
01m   USAPhO 2013, problem B2.

   Remark
   It’s unlikely that you’ll see any examples besides the ones in the above two problems, because
   in almost all other setups, the Poynting vector depends sensitively on the fringe fields, which
   are very hard to calculate. (For some work in this direction, see Energy transfer in electrical
   circuits: A qualitative account.) In any case, the examples above illustrate the important point
   that the energy of a circuit does not flow along the wire, carried by the charges; instead it
   flows into circuit elements from the sides. This was an important early clue of the importance
   of the electromagnetic field.

   Remark
   As proven in Poynting’s theorem, the Poynting vector indeed tells us about the net flow of
   energy. However, this would remain true if we added a constant vector to it, or more generally
   any divergence-free vector field, since these wouldn’t change the net flow. So which option
   is the “correct” one? According to everything we’ve learned so far, there’s no absolute way
   to choose, and we just use the Poynting vector because it’s the simplest option. However, in
   general relativity, the flow of energy directly influences the curvature of spacetime, so there
   is an unambiguous correct answer, which is indeed the Poynting vector.

   Example 3

   Consider two charges q, at positions rx̂ and rŷ respectively, both moving with speed v towards
   the origin. Show that the magnetic forces between them are not equal and opposite. That
   is, electromagnetic forces do not obey Newton’s third law.

   Solution
   In order to find the B field produced by each charge at the location of the other, we use the
   Biot–Savart law and the right-hand rule. Then we use the Lorentz force and the right-hand
   rule again to find the magnetic forces on each charge.

   For example, the B field produced by the first charge at the location of the second is along
   −ẑ. Then the magnetic force on the second charge is parallel to x̂. The magnetic force on
   the first charge is parallel to ŷ. And the force are definitely nonzero, so they can’t be equal
   and opposite.

To explain this, we recall that the point of Newton’s third law is just momentum conservation. This
still holds, as long as one remembers that the field carries momentum of its own. (If we want to
save some version of Newton’s third law, we could say that the real action-reaction pairs are the
forces between the charges and the field, not the charges with each other. But the real lesson is
that Newton’s third law is not fundamental, momentum conservation is.)

       Idea 7
       The momentum density of the electromagnetic field is
                                                         S
                                                 p=         .
                                                         c2
       In other words, momentum density and energy flux density are just proportional. As you
       will see in R2, this is true in general in relativity. The angular momentum density is r × p.
       For an explicit derivation that these definitions ensure the total momentum and angular
       momentum are conserved, see section 8.2 of Griffiths. (You might think the definitions come
       out of nowhere; the straightforward way to find them is to apply Noether’s theorem, as you
       will learn in a more advanced class.)

       Remark
       We have already seen an example of electromagnetic field momentum at work. Back in E4,
       you found that in the presence of a magnetic monopole, the mechanical angular momentum
       L of a point charge was not conserved, but L − qgr̂ was. In fact, this second term turns
       out to be exactly the angular momentum of the field, so this conservation law is simply the
       conservation of total angular momentum. (If you’d like to verify this explicitly, it’s easiest
       to use spherical coordinates with the monopole at the origin and the charge along the z-axis,
       but be warned, it’s fairly messy.)

[3] Problem 14 (Griffiths). A long coaxial cable of length ℓ consists of an inner conductor of radius
    a and an outer conductor of radius b. The inner conductor carries a uniform charge per unit length
    λ, and a steady current I to the right; the outer conductor has the opposite charge and current.

      (a) Find the electromagnetic momentum stored in the fields.

     (b) In part (a) you should have found that the fields contain a nonzero momentum directed along
         the cable. However, this is puzzling because it appears that no net mass is transported along
         the cable. How is this paradox resolved? (Hint: it doesn’t make sense to consider the cable in
         isolation, as nothing would be keeping the current going. Consider attaching a battery across
         the left end and a resistor across the right end.)

Solution:
(a) Set up the obvious cylindrical coordinates, with ẑ directed to the right. In between
         the tubes the fields are
                                                  1 λ             µ0 I
                                           E=            ŝ, B =        ϕ̂,
                                                 2πϵ0 s           2π s
                                                                               R
         and they are zero everywhere else. Therefore, the momentum is ϵ0 (E × B) dV or
                                         µ0 Iλ b 1
                                               Z
                                                                 µ0 Iλℓ
                                  p = ẑ     2      2
                                                      ℓ 2πs ds =         log(b/a) ẑ.
                                         4π     a s               2π
     (b) As this process goes on, the battery loses energy and the resistor gains energy (i.e. heats
         up). Since E = mc2 , that means the resistor is gaining mass while the battery is losing mass.
         Thus, the momentum reflects the fact that the center of mass of the system is going to the
         right. (This is a concrete example of the statement of idea 7, i.e. that momentum is always
         accompanied by the flow of energy.)
          You might also wonder about the momentum and kinetic energy carried by the electrons
          themselves; however, as described in a remark at the end of E5, for typical circuits this is
          negligible compared to the momentum and energy carried by the fields.


[User]
Juku has 12 identical flashlight’s bulbs and a battery. The voltage on the battery’s leads is exactly 5 times bigger than the bulb’s nominal voltage. Juku also found a resistor which has a resistance two times smaller than the resistance of the bulb’s filament in working regimen (the latter was found by dividing the nominal voltage written on the bulb’s socle by the nominal current written on it).\\
\osa How must be the mentioned components connected to a circuit diagram so that all the 12 light bulbs operate with nominal brightness?\\
\osa How many times will the total power of the lights increase (or decrease) if one of the bulbs fuses? Do not account for the relation between the resistance of the lamps and the temperature.

Solution:
\osa Ilmselt tuleb takisti voolu piiramiseks ühendada järjestikku vooluallikaga (ükski teine kombinatsioon, sh takisti ärajätmine, ei võimalda kõiki 12 lampi lülitada nominaalpingele). Lambid saab põhimõtteliselt ahelasse ühendada 1-, 2-, 3-, 4-, 6- või 12-kaupa jadamisi ja seejärel rööbiti (st kõik lambid on ahelasse lülitatud ühetaoliselt). Iga kombinatsiooni korral arvutame esmalt ahela erinevate osade takistuse (ühikutes $R$, mis on üksiku lambi takistus) ja seejärel arvestame pinge jagunemist elementidel proportsionaalselt takistusega. Erinevate variantide kontrollimine näitab, et lambile langeb nominaalpinge näiteks juhul, kui ühendada lambid kahekaupa jadamisi ja seejärel 6 sellist ahelat rööbiti. Seega, õige skeem on selline, nagu kujutatud joonisel.

\begin{center}
\includegraphics{2018-v2g-08-lambid-joonis.pdf}
\end{center}

Kuni kõik lambid põlevad, langeb igale lambile pinge $U/5$, kus $U$ on pingeallika klemmipinge. Koguvõimsus on vastavalt
\[
P_1=12\times \frac{U^2}{25R}=\frac{12U^2}{25R},
\]
kus $R$ on üksiku lambi takistus. Kui üks lampidest põleb läbi, siis ka sellega järjestikku ühendatud pirn lõpetab töötamise. Järele jääb vaid 5 rööbiti ühendatud ahelat ja seega lampide kogutakistus on $2R/5$. Võttes nüüd arvesse pinge jagunemist ahelas, saab pinge igal lambil olema
\[
U\times \frac{2R/5}{2R/5+R/2} \times \frac{1}{2}=\frac{2}{9}U
\]
ja lampide koguvõimsus vastavalt
\[
P_2=10\times \frac{4U^2}{81R}=\frac{40U^2}{81R}.
\]
Seega koguvõimsus kasvab $P_2/P_1=250/243\approx \num{1.029}$ korda.

\textit{Märkus.} Leidub veel mitu sobivat skeemi. Toome mõned neist.
\begin{itemize}
\item
Ühendame lambid kolmekaupa jadamisi ja seejärel 4 sellist ahelat rööbiti. Sel juhul osas \textit{b)} kahaneb lampide koguvõimsus $\num{1.080}$ korda.
\item
Ühendame lambid neljakaupa rööbiti ja seejärel 3 sellist plokki jadamisi takistiga. Sel juhul osas \textit{b)} kahaneb lampide koguvõimsus $\num{1.024}$ korda
\end{itemize}
Töötavaid skeeme leidub rohkemgi.

12 lampi

Juku käsutuses on 12 ühesugust taskulambipirni ning patarei, mille klemmipinge on täpselt 5 korda suurem pirni nimipingest. Lisaks leidis ta juhtumisi takisti, mille takistus on parajasti pool lambi hõõgniidi takistusest töörežiimis (viimase sai ta teada jagades lambi soklile kirjutatud nimipinge ja -voolu omavahel).\\
\osa Kuidas tuleb ühendada nimetatud komponendid elektriahelasse, et kõik 12 pirni põleksid normaalheledusega?\\
\osa Mitu korda kasvab (või kahaneb) lampide koguvõimsus, kui üks lampidest läbi põleb? Lampide takistuse sõltuvust temperatuurist võib jätta arvestamata.

Eesmärk on leida elektriskeem, kus igale lambile langeb viiendik klemmipingest. Mõistlik on kõigepealt proovida võimalikult lihtsaid korrapäraseid lampide konfiguratsioone. Näiteks saab lambid asetada rööpühendusse, kus igas harus on kas 1, 2, 3, 4, 6 või 12 lampi.


[User]
Venimatu ja kaalutu niidi otsa kinnitati kuulike.
Niit viidi horisontaalasendisse ja lasti lahti. Kuulikese kiiruse vertikaalne komponent
hakkab esialgu suurenema, kuid teatud hetkest alates vähenema. Millise nurga moodustab niit vertikaalsihiga ajahetkel, kui kuulikese kiiruse vertikaalne komponent on
maksimaalne?

Solution:
Olgu $\alpha$ nurk vertikaali ja varda vahel ning l niidi pikkus. Energia jäävuse seadusest
$v^2 = 2gl \cos \alpha$, millest kiiruse vertikaalkomponendi ruut
\[
v_{y}^{2}=2 g l \cos \alpha \sin ^{2} \alpha=2 g l \cos \alpha\left(1-\cos ^{2} \alpha\right).
\]
Tähistades $\cos \alpha = y$, saame $v_y^2$ maksimumi tingimuse, kui võtame sellest tuletise $y$ järgi:
\[
\frac{\D v_{y}^{2}}{\D y}=2 g l\left( 1-3 y^{2}\right)=0,
\]
millest $y = 1/ \sqrt 3$ ja $\alpha = \arccos \left( 1/ \sqrt 3\right) \approx \ang{55}$.

\vspace{0.5\baselineskip}

\begin{wrapfigure}{r}{0.3\textwidth}
	\begin{center}
		\includegraphics[width=0.95\linewidth]{2008-lahg-05-lah}
	\end{center}
\end{wrapfigure}
\emph{Alternatiivne lahendus}\\
Ülesandes on öeldud, et kuulikese kiiruse vertikaalne komponent
hakkab esialgu suurenema, kuid teatud hetkest alates vähenema.
See tähendab seda, et kuulikese kiirenduse vertikaalne komponent oli alguses positiivne ning pärast muutus negatiivseks. Järelikult hetkel, kui kiiruse vertikaalne komponent on maksimaalne, on kiirenduse vertikaalne komponent null. Seega vaadeldaval
hetkel võrdub kuulikese raskusjõu ja niidi tõmbejõu vertikaalsete
projektsioonide summa nulliga ehk siis
\begin{equation} \label{2008-lahg-05:eq1}
mg = T \cos \alpha.
\end{equation}
Kuuli kesktõmbekiirendus on $a_n = v^2/l$, kus $l$ on niidi pikkus. Seega Newtoni II
seaduse põhjal
\begin{equation} \label{2008-lahg-05:eq2}
m\frac{v^2}{l} = T - mg \cos \alpha.
\end{equation}
Energia jäävuse seadusest
\begin{equation} \label{2008-lahg-05:eq3}
v^2 = 2gl \cos \alpha.
\end{equation}

Asendades võrdusesse (\ref{2008-lahg-05:eq2}) avaldised $T$ ja $v^2$
jaoks võrdustest (\ref{2008-lahg-05:eq1}) ja (\ref{2008-lahg-05:eq3}) saame
\[
\frac{m \cdot 2 g l \cos \alpha}{l}=\frac{m g}{\cos \alpha}-m g \cos \alpha,
\]
millest
\[
\frac{1}{\cos \alpha}-\cos \alpha=2 \cos \alpha \quad \Rightarrow\quad \cos \alpha=\frac{1}{\sqrt{3}}.
\]
Siit $\alpha = \arccos \left( 1/ \sqrt 3\right) \approx \ang{55}$.

Kuulike

Venimatu ja kaalutu niidi otsa kinnitati kuulike.
Niit viidi horisontaalasendisse ja lasti lahti. Kuulikese kiiruse vertikaalne komponent
hakkab esialgu suurenema, kuid teatud hetkest alates vähenema. Millise nurga moodustab niit vertikaalsihiga ajahetkel, kui kuulikese kiiruse vertikaalne komponent on
maksimaalne?

Kuulikese kiirus on leitav energia jäävuse seadusest. Edasi taandub ülesanne ekstreemumpunkti leidmisele.


[User]
Question A1
The Doppler effect for a source moving relative to a stationary observer is described by
                                                               f0
                                                    f=
                                                         1 − (v/c) cos θ

where f is the frequency measured by the observer, f0 is the frequency emitted by the source, v is
the speed of the source, c is the wave speed, and θ is the angle between the source velocity and the
line between the source and observer. (Thus θ = 0 when the source is moving directly towards the
observer and θ = π when moving directly away.)
    A sound source of constant frequency travels at a constant velocity past an observer, and the
observed frequency is plotted as a function of time:

                   450
                                                                                Observed Frequency
                   448
                   446
                   444
                   442
                   440

  Frequency (Hz)
                   438
                   436
                   434
                   432
                   430
                   428
                   426
                   424
                   422
                   420
                         0   1      2    3     4     5     6      7     8   9   10    11     12   13   14
                                                               Time (s)

       The experiment happens in room temperature air, so the speed of sound is 340 m/s.

  a. What is the speed of the source?

  b. What is the smallest distance between the source and the observer?

                                 Copyright c 2016 American Association of Physics Teachers

2016 USA Physics Olympiad Exam                      Part A                                              4

Solution:
Question A1
The Doppler effect for a source moving relative to a stationary observer is described by
                                                               f0
                                                    f=
                                                         1 − (v/c) cos θ

where f is the frequency measured by the observer, f0 is the frequency emitted by the source, v is
the speed of the source, c is the wave speed, and θ is the angle between the source velocity and the
line between the source and observer. (Thus θ = 0 when the source is moving directly towards the
observer and θ = π when moving directly away.)
    A sound source of constant frequency travels at a constant velocity past an observer, and the
observed frequency is plotted as a function of time:

                   450
                                                                                Observed Frequency
                   448
                   446
                   444
                   442
                   440

  Frequency (Hz)
                   438
                   436
                   434
                   432
                   430
                   428
                   426
                   424
                   422
                   420
                         0   1      2    3     4     5     6      7     8   9   10    11     12   13   14
                                                               Time (s)

       The experiment happens in room temperature air, so the speed of sound is 340 m/s.

  a. What is the speed of the source?

                                 Copyright c 2016 American Association of Physics Teachers

2016 USA Physics Olympiad Exam               Part A                                                4

                                              Solution

    For θ = 0 we have
                                           fa ≈ f0 /(1 − v/c)
    and for θ = π,
                                           fb = f0 /(1 + v/c).
    Read fa and fb off the early and late time portions of the graph and use

                                      fa /fb = (1 + v/c)/(1 − v/c)

    giving an answer of v = 10.7 m/s.
    Alternatively, we can see that v  c and approximate

                                           fa /fb ≈ 1 + 2v/c

    which makes the calculation of v slightly faster. This is acceptable because the error terms
    are of order (v/c)2 ∼ 0.1%.

  b. What is the smallest distance between the source and the observer?

                                              Solution

    Let d be the (fixed) distance between the observer and the path of the source; let x be the
    displacement along the path, with x = 0 at closest approach. Then for |x|  d,

                                          cos θ ≈ cot θ = x/d

    so we have
                            f = f0 /(1 − (v/c)(x/d)) ≈ f0 (1 + (v/c)(x/d)).
    Taking the time derivative, and noting that x0 is simply v,

                                            f 0 = f0 (v 2 /c)d

    Therefore we can read f 0 off the center region of the graph. We still need to find f0 , which we
    can do using our result from part (a) or simply by averaging fa and fb , since v  c, giving
    f0 = 435 Hz and an answer of d = 17.8 m.
    There’s also a nice trick to speed up this computation. Draw lines at the asymptotic values
    and through the central data points. The two horizontal lines are 2f0 (v/c) apart in frequency,
    so the time between their intersections with the third line is simply 2d/v.

                 Copyright c 2016 American Association of Physics Teachers

2016 USA Physics Olympiad Exam                   Part A                                                 5

USAPhO 2016 Question A1


[User]
A uniform half-disc of radius R is nailed to a wall at the center of its circle and
    allowed to come to equilibrium. The half-disc is then rotated by an angle dθ. By calculating the
    energy needed to do this in two different ways, find the distance from the pivot point to the center
    of mass.
[4] Problem 25 (Morin 5.31). Assume that a cloud consists of tiny water droplets suspended (uniformly
    distributed, and at rest) in air, and consider a raindrop falling through them. Assume the raindrop is
    initially of negligible size, remains spherical at all times, and collides perfectly inelastically with the
    droplets. It turns out that the raindrop accelerates uniformly; assuming this, find the acceleration.
[3] Problem 26 (Kvant). Half of a flexible pearl necklace lies on a horizontal frictionless table, while
    the other half hangs down vertically at the edge. If the necklace is released from rest, it will slide off
    the table. At some point, the hanging part of the necklace will begin to whip back and forth. What
    fraction of the necklace is on the table when this begins? (Hint: we are considering a pearl necklace
    with no empty string between adjacent pearls; as a result, all the pearls accelerate smoothly. To
    solve the problem, think about the vertical forces. There is an important related problem in M2.)
[4] Problem 27 (BAUPC 2002). A small ball is attached to a massless string of length L, the other
    end of which is attached to a very thin pole. The ball is thrown so that it initially travels in a
    horizontal circle, with the string making an angle θ0 with the vertical. As time goes on, the string
    wraps itself around the pole. Assume that (1) the pole is thin enough so that the length of string in
    the air decreases very slowly, and (2) the pole has enough friction so that the string does not slide
    on the pole, once it touches it. Show that the ratio of the ball’s final speed (right before it hits the
    pole) to initial speed is sin θ0 .
    When dealing with an extended system whose parts all move in different ways, conservation of
    energy is occasionally useless. However, the somewhat obscure idea of “center of mass energy” may
    become useful instead. For more about this concept, see section 13.5 of Halliday and Resnick.

        Idea 7: Center of Mass Energy

        The work done on a part of a system is

                                                   dW = F dx

        where F is the force on that specific part of the system, and dx is its displacement. Then
        dW = dE where E is the total energy of the system.

        Similarly, the “center of mass work” done on a system is

                                                 dWcm = F dxcm

        where F is the total force on the system and dxcm is the displacement of the center of mass.
        Then dWcm = dEcm where the “center of mass energy” is defined as Ecm = M vcm   2 /2.

       It should be noted that, like regular energy and work, center of mass energy and work depend
       on the reference frame you’re using.

       Example 11

       Consider a cyclist who pedals their bike to accelerate. The wheels roll without slipping on
       the ground. The cyclist moves a distance d, with the bike experiencing a constant friction
       force f from the ground. Analyze the situation using both energy and center of mass energy.

       Solution
       Since the wheels roll without slipping, their contact point with the ground is always zero,
       so the friction force does exactly zero work. Thus the net energy of the cyclist/bike system
       is conserved. The additional kinetic energy of the cyclist/bike comes from the chemical
       energy of the cyclist, which ultimately came from what they ate. So conservation of energy
       is correct, but it doesn’t tell us anything useful at all.

       Now consider center of mass energy. Considering the cyclist/bike system, the center of mass
                                              2 /2. This allows us to compute the change in velocity
       work is f d, which is the change in M vcm
       of the cyclist/bike.

       Example 12

       Consider the same setup as in the previous example, but now the cyclist brakes hard. The
       wheels slip on the ground, and experience a friction force −f while the cyclist moves a
       distance d. Analyze the situation using both energy and center of mass energy.

       Solution
       The center of mass work equation tells us about the overall deceleration of the cyclist/bike,
       just as in the previous example.

       On the other hand, the work done by the friction force is indeterminate! It can be any
       quantity between zero and −f d. When it is 0, the total energy of the cyclist/bike system
       is again conserved, which means all the kinetic energy lost is dissipated as heat inside the
       bike itself. When it is −f d, all the kinetic energy lost is dissipated as heat in the ground,
       and hence energy is removed from the cyclist/bike system. In general, the work will be
       an intermediate value, meaning that both the ground and the bike heat up, but we can’t
       calculate what it is without a microscopic model of how the friction works. It depends on,
       e.g. how easily the ground and bike tire surface deform.

Solution:
Refer to the diagram below.

    Suppose the center of mass is at radius r. The energy required to turn the disc by dθ is
                                                                mgx 2
                                         (1 − cos(dθ)) mgx ≈       dθ .

    However, when rotated, all that has changed is that there is a new sector of angle dθ above (sector
    A), and one sector is now missing (sector B). A thin sector is approximately an isosceles triangle, so
    its center of mass is at radius 2R/3. Thus, the total extra energy of A is (dm)g(2R/3)dθ/2, where
    dm/m = dθ/π, so it’s mgR(1/3π) dθ2 , and the extra energy due to the absence of B is the same,
    so we have
                                           2              mgx 2
                                              mgR dθ2 =        dθ
                                          3π                2
    which implies

                                                 x=      R.
                                                      3π
[4] Problem 25 (Morin 5.31). Assume that a cloud consists of tiny water droplets suspended (uniformly
    distributed, and at rest) in air, and consider a raindrop falling through them. Assume the raindrop is
    initially of negligible size, remains spherical at all times, and collides perfectly inelastically with the
    droplets. It turns out that the raindrop accelerates uniformly; assuming this, find the acceleration.

    Solution. Suppose the mass density in the cloud is λ and the mass density of the raindrop is ρ
    (note ρ > λ), and suppose r is the radius of the drop, M the mass, and v the velocity. We see that

                                                                   ṙ
                                              Ṁ = 4πr2 ṙρ = 3M
                                                                   r
    and
                                                   Ṁ = πr2 vλ,

    which combine to give
                                                           4ρ
                                                    v=        ṙ.
                                                           λ
    We see that M g = Ṁ v + M v̇, so Newton’s second law is

                                                   3ṙ 4ρ     4ρ
                                        Mg = M            ṙ + r̈
                                                   r λ        λ
    and writing everything in terms of r gives

                                              rgλ/ρ = 12ṙ2 + 4rr̈.

    This is a nonlinear second-order differential equation; there is no general method to solve these
    equations. Certainly an exponential won’t work, because you won’t get the same exponential on
    the left and right-hand sides. However, we can use the hint, which indicates that v is linear in time.
    This implies that r is a quadratic, so guessing r = At2 gives

                                          At2 gλ/ρ = 4A2 (12t2 + 2t2 ).

    This implies that we indeed have a solution, as long as

                                                A = (gλ/ρ)/56.

    Using our relation between v and ṙ, we finally have
                                                         4ρ      g
                                                  v̇ =      r̈ =
                                                         λ       7
    which is the acceleration.
       As you can see from this problem and an earlier one, nonlinear second-order differential equations
    are actually quite common in physics. Trying a pure power Atn is a decent first guess, because
    monomials remain monomials under both differentiation and multiplication; for the same reason,
    an exponential AeBt can also work. However, in practice, the vast majority of such differential
    equations don’t have analytic solutions at all, or only have solutions in terms of exotic special
    functions. Problems for Olympiads and undergraduate textbooks are generally chosen precisely to
    avoid these complications, since they draw attention away from the essential physics.
       This raindrop problem is a classic, first invented (though with a slightly different mass accretion
    rule) for the infamous Cambridge Smith’s Prize Examination in 1853. Many papers have been
    written about it, such as this one by K. S. Krane, who is the K in HRK.

[3] Problem 26 (Kvant). Half of a flexible pearl necklace lies on a horizontal frictionless table, while
    the other half hangs down vertically at the edge. If the necklace is released from rest, it will slide off
    the table. At some point, the hanging part of the necklace will begin to whip back and forth. What
    fraction of the necklace is on the table when this begins? (Hint: we are considering a pearl necklace
    with no empty string between adjacent pearls; as a result, all the pearls accelerate smoothly. To
    solve the problem, think about the vertical forces. There is an important related problem in M2.)

    Solution. Physically, what happens is that a sizable force is needed to turn the pearls around when
    they reach the corner, to go from moving horizontally to vertically. At a certain critical velocity,
    tension will no longer be enough to do this, and the pearl necklace will jump off the corner. This
    will lead to the hanging part whipping back and forth.

        To see when this happens, consider the vertical forces on the pearl necklace. Suppose the necklace
    has mass M and a fraction x is hanging off the table. As we saw with a similar but static problem
    in M2, the normal force on the horizontal part of the table has to cancel the force of gravity acting
    on the part of the necklace on the table. This is because otherwise a piece of the horizontal part of
    the necklace would have an unbalanced vertical force, and would have to go into the table or jump
    off it, neither of which make sense.
        Thus, considering the vertical forces just gives
                                              dpy
                                                  = xM g − Nc,y
                                               dt
    where we take the downward direction as positive for convenience, and Nc,y is the vertical part of
    the normal force acting at the corner of the table. (Recall from M2 that this is significant even
    when the corner is small.) Since the pearl necklace is flexible, the process is elastic, so energy is
    conserved. Our strategy is to use energy conservation to find dpy /dt and use that to find Nc,y . The
    necklace jumps off the corner when Nc,y becomes zero.
       Taking the necklace to have length L, energy conservation gives
               1              x M gL                              p   p
                 M v 2 = M xgL −     =⇒ v 2 = gL(x2 − 1/4) =⇒ ẋ = g/L x2 − 1/4.
               2              2  8
    The acceleration is
                       p       1              p        x      p   p
                ẍ =    g/L p         · 2xẋ = g/L p           g/L x2 − 1/4 = (g/L)x
                           2 x2 − 1/4                x2 − 1/4

    which we also found using the generalized coordinate approach explained in problem 16. Thus,
         dpy      d
             = M L (xẋ) = M L(xẍ + ẋ2 ) = M L((g/L)x2 + (g/L)(x2 − 1/4)) = M g(2x2 − 1/4).
          dt      dt
    Setting Nc,y to zero gives
                                                  2x2 − 1/4 = x
                                              √
    for which the relevant root is x = (1 +    3)/4. At this point, the fraction on the table is
                                                           √
                                                       3− 3
                                               1−x=            .

    This is barely less than 1/2, so the necklace jumps almost immediately!
       Ideal flexible ropes can have rather unintuitive behavior, and we will discuss them further in a
    problem in W1. By the way, you might   p have noticed that the necklace jumps precisely when its
    speed is equal to the wave speed v = T /(M/L) at the corner; the explanation for this will also
    be found in W1.

[4] Problem 27 (BAUPC 2002). A small ball is attached to a massless string of length L, the other
    end of which is attached to a very thin pole. The ball is thrown so that it initially travels in a
    horizontal circle, with the string making an angle θ0 with the vertical. As time goes on, the string
    wraps itself around the pole. Assume that (1) the pole is thin enough so that the length of string in
    the air decreases very slowly, and (2) the pole has enough friction so that the string does not slide
    on the pole, once it touches it. Show that the ratio of the ball’s final speed (right before it hits the
    pole) to initial speed is sin θ0 .

Solution. The official solution is here, but it’s a lot longer because it also solves for the evolution
of the height of the ball; in this shorter version of the problem, we only want the final speed, which
is a lot easier to find.
    In this problem, energy is conserved because the string doesn’t slide, but angular momentum isn’t
conserved. Now note that if θ were constant, the ball would move at a constant height. Changes in θ
as the string winds up change the ball’s height, which then changes its speed by energy conservation.
By doing some geometry, you can show

                                       mv dv = −mgℓ sin θ dθ.

The radial F = ma equation is
                                                        mv 2
                                          mg tan θ =           .
                                                       ℓ sin θ
Substituting this into the equation above gives
                                             mv 2
                                      −mg            dθ = mv dv
                                            mg tan θ
which upon separation gives the simple result
                                     Z θf            Z vf
                                                          dv
                                   −      cot θ dθ =         .
                                       θ0             v0  v
Therefore, we have
                                             sin θ0  vf
                                                    = .
                                             sin θf  v0
Now we just need to know the final angle θf . If we substitute this back into the radial F = ma
equation, then at the end of the motion we have tan θf ∝ vf2 /(ℓ sin θf ) ∝ 1/(ℓ sin3 θf ) with ℓ → 0.
Since sin θf is finite, this is only possible if tan θf diverges, so θf = π/2. We therefore conclude the
desired result, vf /v0 = sin θ0 .
When dealing with an extended system whose parts all move in different ways, conservation of
energy is occasionally useless. However, the somewhat obscure idea of “center of mass energy” may
become useful instead. For more about this concept, see section 13.5 of Halliday and Resnick.

   Idea 7: Center of Mass Energy

   The work done on a part of a system is

                                             dW = F dx

   where F is the force on that specific part of the system, and dx is its displacement. Then
   dW = dE where E is the total energy of the system.

   Similarly, the “center of mass work” done on a system is

                                            dWcm = F dxcm

   where F is the total force on the system and dxcm is the displacement of the center of mass.
   Then dWcm = dEcm where the “center of mass energy” is defined as Ecm = M vcm   2 /2.

       It should be noted that, like regular energy and work, center of mass energy and work depend
       on the reference frame you’re using.

       Example 11

       Consider a cyclist who pedals their bike to accelerate. The wheels roll without slipping on
       the ground. The cyclist moves a distance d, with the bike experiencing a constant friction
       force f from the ground. Analyze the situation using both energy and center of mass energy.

       Solution
       Since the wheels roll without slipping, their contact point with the ground is always zero,
       so the friction force does exactly zero work. Thus the net energy of the cyclist/bike system
       is conserved. The additional kinetic energy of the cyclist/bike comes from the chemical
       energy of the cyclist, which ultimately came from what they ate. So conservation of energy
       is correct, but it doesn’t tell us anything useful at all.

       Now consider center of mass energy. Considering the cyclist/bike system, the center of mass
                                              2 /2. This allows us to compute the change in velocity
       work is f d, which is the change in M vcm
       of the cyclist/bike.

       Example 12

       Consider the same setup as in the previous example, but now the cyclist brakes hard. The
       wheels slip on the ground, and experience a friction force −f while the cyclist moves a
       distance d. Analyze the situation using both energy and center of mass energy.

       Solution
       The center of mass work equation tells us about the overall deceleration of the cyclist/bike,
       just as in the previous example.

       On the other hand, the work done by the friction force is indeterminate! It can be any
       quantity between zero and −f d. When it is 0, the total energy of the cyclist/bike system
       is again conserved, which means all the kinetic energy lost is dissipated as heat inside the
       bike itself. When it is −f d, all the kinetic energy lost is dissipated as heat in the ground,
       and hence energy is removed from the cyclist/bike system. In general, the work will be
       an intermediate value, meaning that both the ground and the bike heat up, but we can’t
       calculate what it is without a microscopic model of how the friction works. It depends on,
       e.g. how easily the ground and bike tire surface deform.


[User]
Eiffeli torni ülemiselt vaateplatvormilt (kõrgus maapinnast
$h=\SI{273}{m}$) lastakse kukkuda raudkuulil. Täpselt $t=3$ sekundi pärast kukutatakse
veel üks raudkuul. Kui suur on raudkuulide suurim
kiiruste vahe langemisel? Kui suur on ajavahemik kuulide maapinnale jõudmiste
vahel? Raskuskiirendus $g=\SI{9.8}{m/s^2}$. Katse käigus ükski külastaja viga ei saanud.

Solution:
Kontrollime, kui kaua kukub raudkuul ülemiselt vaateplatvormilt $h=\SI{273}{m}$ maapinnale.
\[h=\frac{gt^2}{2} \Rightarrow t=\sqrt{\frac{2h}{g}}\approx \SI{7.5}{s}.\]
Alates hetkest, kui mõlemad kuulid langevad, on nende suhteline kiirus muutumatu, sest mõlemad kuulid on siis vabalt langevas taustsüsteemis paigal.
Leiame esimese kuuli kiiruse teise kuuli kukutamise hetkel.
\[v=gt\approx \SI{29.4}{m/s}.\]
Ajavahemik kuulide maapinnale jõudmisel on sama mis kuulide kukutamiselgi ehk $t=\SI{3}{s}$.

Eiffeli torn

Eiffeli torni ülemiselt vaateplatvormilt (kõrgus maapinnast
$h=\SI{273}{m}$) lastakse kukkuda raudkuulil. Täpselt $t=3$ sekundi pärast kukutatakse
veel üks raudkuul. Kui suur on raudkuulide suurim
kiiruste vahe langemisel? Kui suur on ajavahemik kuulide maapinnale jõudmiste
vahel? Raskuskiirendus $g=\SI{9.8}{m/s^2}$. Katse käigus ükski külastaja viga ei saanud.

Kukkumise käigus kiirenevad mõlemad kuulid sama kiirusega. Seega on nende suhteline kiirus muutumatu.


[User]
Veepargis on punane ja roheline liumägi. Joonisel on näidatud nende profiilid. Mari alustab punaselt liumäelt laskumist algkiiruseta, Juri aga lasküb roheliselt mäelt algkiirusega $v_0 = \SI{1}{m/s}$. Kumma kiirus on laskumise lõpuks suurem ja mitu korda, kui Jüri kaalub $m_j = \SI{70}{kg}$ ja Mari $m_m = \SI{60}{kg}$? Võib eeldada, et laskumise ajal mõjub neile mõlemale konstantne takistusjõud $F = \SI{20}{N}$.
\begin{center}
	\includegraphics[width=0.7\linewidth]{2004-v3g-02-yl.pdf}
\end{center}

Solution:
Nii Jüri kui ka Mari puhul peab kehtima energia jäävuse seadus:
$$
\frac{1}{2} m v_{0}+m g h=\frac{1}{2} m v+F s,
$$
kus $v_{0}$ on algkiirus, $h-$ mäe korgus, $v$ - lõppkiirus ja $F s-$ töö takistusjõu vastu. Avaldis lõppkiiruse jaoks saab kuju
$$
v=\sqrt{v_{0}^{2}+2 g h-2 F s / m}.
$$
Jooniselt leiame, et $h=\SI{7}{m}$. $s$ määramiseks leiame teepikkuse iga sirge lõigu jaoks ning liidame kokku. Jüri läbis teepikkuse
$$
s_{j}=\sqrt{20^{2}+4^{2}}+\sqrt{35^{2}+3^{2}}=\SI{55,5}{m}
$$
ning Mari teepikkus oli
$$
s_{m}=\sqrt{25^{2}+3^{2}}+15+\sqrt{30^{2}+4^{2}}=\SI{70,4}{m}
$$
Lõppkiiruse väärtuseks saame nüüd Mari jaoks
$$
v_{m}=\sqrt{0+2 \cdot 9,8 \cdot 7-\frac{2 \cdot 20 \cdot \num{70,4}}{60}} \approx \SI{9,5}{m/s}
$$
ning Jüri jaoks
$$
v_{j}=\sqrt{1+2 \cdot 9,8 \cdot 7-\frac{2 \cdot 20 \cdot \num{55,5}}{70}} \approx \SI{10,3}{m/s}
$$
Seega Jüri on lõpus \num{1,08} korda kiirem kui Mari.

Veepark

Veepargis on punane ja roheline liumägi. Joonisel on näidatud nende profiilid. Mari alustab punaselt liumäelt laskumist algkiiruseta, Juri aga lasküb roheliselt mäelt algkiirusega $v_0 = \SI{1}{m/s}$. Kumma kiirus on laskumise lõpuks suurem ja mitu korda, kui Jüri kaalub $m_j = \SI{70}{kg}$ ja Mari $m_m = \SI{60}{kg}$? Võib eeldada, et laskumise ajal mõjub neile mõlemale konstantne takistusjõud $F = \SI{20}{N}$.
\begin{center}
	\includegraphics[width=0.7\linewidth]{2004-v3g-02-yl.pdf}
\end{center}


[User]
10. YOUNG’S MODULUS OF RUBBER (12 points)

The linear Hooke’s law for a
rope made from an elastic material is sup-
posed to held for small relative deformations
ε= x/L (which is also called “strain”), where
L is the undeformed length of the rope, and x
is the deformation. Once ε becomes too large,
the force-deformation relationship F = kx is
no longer linear; what is “too large” depends
on the material. For very elastic materials
which can reach relative deformations con-
siderably large than one, it may happen that
the linear Hooke’s law with a constant stiff-
ness k fails, but if we take into account the
change of the cross-sectional area S of the
rope with k = ES/L, where E is the Young’s
modulus of the elastic material, such a non-
linear Hooke’s law remains valid. In that
case we can say that there is still a linear
stress-strain relationship σ= E·ε, where the
stress σ= F/S.
i) (7 points) Measure the relationship
between the stress σ and strain ε in a rubber
string and plot it.
ii) (5 points) From your plot determine the
Young’s modulus E with its uncertainty, and
the maximum strain εm until which it ap-
plies.
Note: the diameter of the thread is to be
measured using the diffraction of laser light.
Equipment: rubber thread, stand, meas-
uring tape, 15 hex nuts with a known mass,
a plastic bag for hanging a set of nuts to the
thread, a green laser (λ=532nm), a screen.
WARNING: AVOID LOOKING INTO
A LASER BEAM, THIS MAY DAMAGE
YOUR EYES!

Solution:
10. Young's modulus of rubber

1) The setup is as follows. The rubber thread
is fixed to the stand, and the plastic bag is fixed
to the free end of the thread. The hex nuts are
added, one-by-one, into the bag, starting with
zero and ending with 15. The laser light is dir-
ected to the thread and the diraction pattern
is observed on the screen (which is a vertically
fixed sheet of graphic paper on a support). The
diraction pattern from a wire is the same as
from a single slit (the superposition of the Huy-
gens sources from those two cases gives a full set
of sources on a flat wave front, hence the electric
fields from those two cases must provide equal
and opposite wave amplitudes and equal intens-
ities). So, if we measure on the screen the dis-
tance a between such maxima which are separ-
ated by n (e.g. n = 10) periods of the diraction
pattern then using small-angle-approximation,
nλ/d = a/L, where L is the distance from the
thread to the screen and d  the thread dia-
meter. So,
d = nLλ/a.
The strain is calculated by making markings on
the thread and measuring the distance b between
these in a stretched state,
ε =
b − b0
b0
,
where b0 is the length in a non-strained state.
The stress is calculated as
σ =
4Nmg
πd2
,
where N is the number of hex nuts in the bag
and m  the mass of a single nut. The data
are plotted in a graph; linear relationship cor-
responds to a straight line. The uncertainties
are calculated using the rule of relative uncer-
tainties, either using Pythagorean or simple ad-
dition, e.g.
∆ε = ε∆b

2
b − b0
+
1
b0

,
where ∆b ≈ 0.5mm is the length measurement
uncertainty. Similarly,
∆σ = 2σ
∆d
d
.
These uncertainties are marked in the graph as
error bars.
2) Using the plot, we need to determine such a
value of ε = ε∗ that for ε1 > ε∗, it is impossible
to draw a straight line intersecting the error bars
of all the data points with ε1 < ε∗.

YOUNG’S MODULUS OF RUBBER


[User]
Theoretical Competition-Problem No. 3
           Compression and expansion of a two gases system

     A cylinder is divided in two compartments with a mobile partition NM. The compartment
in the left is limited by the fond of the cylinder and the
partition NM (Figure 1). This compartment contains             V0     N       V0 A
one mole of water vapor. The compartment in the right
is limited by the partition NM and a mobile piston AB.
                                                               p1            p1
This compartment contains one mole of nitrogen gas
                                                               T1           T1
(N2).
     At first, the volumes and temperatures of the gases
in two compartments are equal. The partition NM is                     M           B
well heat conductive. His heat capacity is very small              Figure 1
and can be neglected.
     The specific volume of liquid water is negligible in comparison with the specific volume
of water vapor at the same temperature.
     The specific latent heat of vaporization L is defined as the amount of heat that must be
delivered to one unit of mass of substance to convert it from liquid state to vapor at the same
temperature. For water at T0 = 373 K, L = 2250 kJ/kg.

1. Suppose that the piston and the wall of the cylinder are heat conductive, the partition NM
can slide freely without friction. The initial state of the gases in the cylinder is defined as
follows:
    Pressure p1 = 0.5 atm.; total volume V1 = 2V0; temperature T1 = 373 K.
    The piston AB slowly compresses the gases in a quasi-static (quasi-equilibrium) and
isothermal process to the final volume VF = V0/4
    a. Draw the p(V) curve, that is the curve representing the dependence of pressure p on the
total volume V of both gases in the cylinder at temperature T1. Calculate the coordinates of
important points of the curve. [1.5 pts]
    Gas constant: R = 8.31 J/mol.K or R = 0.0820 L.atm./mol.K
    1 atm. =101.3 kPa;
    Under the pressure p0 =1 atm., water boils at the temperature T0 = 373 K.
    b. Calculate the work done by the piston in the process of gases compressing. [1.0 pts]
           dV
        ∫ V = ln V
    c. Calculate the heat delivered to outside in the process. [1.5 pts]

2. All conditions as in 1. except that there is friction between partition NM and the wall of
the cylinder so that NM displaces only when the difference of the pressures acting on its two
opposed faces attains 0.5 atm. and over (assuming that the coefficients of static and kinetic
friction are equal).
    a. Draw the p(V) curve representing the pressure p in the right compartment as a
function of the total volume V of both gases in the cylinder at a constant temperature T1.
[1.5 pts]
    b. Calculate the work done by the piston in compressing the gases. [0.5 pts]

                                                                                1

    c. After the volume of gases reaches the value VF = V0/4, piston AB displaces slowly to
the right and makes a quasi-static and isothermal process of expansion of both substances
(water and nitrogen) to the initial total volume 2V0. Continue to draw in the diagram in
question 2.a. the curve representing this process [2.0 pts]

    Hint for 2.
    Create a table like the one shown here and use it to draw the curves as required in 2.a.
and 2.c.
State Left compartment             Right compartment         Total       Pressure on
         Volume | Pressure         Volume | Pressure        volume       piston AB
initial V0       | 0.5 atm.        V0       | 0.5 atm.        2V0            0.5 atm.
   2             |                          |
   3             |                          |
   .             |                          |
   .             |                          |
   .             |                          |
   .             |                          |
   .             |                          |
 final           |                          |                 2V0

3. Suppose that the wall and the fond of the cylinder and the piston are heat insulator, the
partition NM is fixed and heat conductive, the initial state of gases is as in 1. Piston AB
moves slowly toward the right side and increases the volume of the right compartment until
the water vapor begins to condense in the left compartment.
    a. Calculate the final volume of the right compartment. [3 pts]
    b. Calculate the work done by the gas in this expansion. [1 pts]
                                                             C                         7
    The ratio of isobaric heat capacity to isochoric one γ = p for nitrogen is γ 1 =      and
                                                             CV                        5
                      8
for water vapor γ 2 = .
                      6
    In the interval of temperature from 353 K to 393 K one can use the following
approximate formula:
                                     µL 1 1 
                         p = p0 exp −    ( − )
                                      R   T T0 
where T is boiling temperature of water under pressure p, µ - its molar mass. p0, L and T0 are
given above.

                                                                                2

Solution:
Solution of Problem No. 3
                  Compression and expansion of a two gases system

1.
     a. The isotherm curve is shown in Figure 1.

                         p[atm.]
                                2

                                1
                              0.5
                               0
                                      V0   V0
                                                      V0             2V0    V
                                      4    2

                                            Figure 1

               RT1      8.31 × 373
        V0 =       =                    = 0.0612 m3
                p1   0.5 × 1.013 × 10 5

                                        = 61.2 dm3

     b. The process of compressing can be divided into 3 stages:

         (p1, 2V0) →       (2p1,V0)    →        (2p1,V0 /2) →   (4p1,V0 )
              (1)               (2)                   (3)           (4)

The work in each stage can be calculated as follows:
             V0               2V0 dV
    A12 = - ∫ pdV = 2RT1 ∫             = 2RT1 ln2 = 4297 J
             2V0             V0     V
    A23 = 2p1(V0- V0/2) = RT1 = 3100 J
              1                1
                V0               V0 dV
    A34 = - ∫1 p ' dV = RT1 ∫12
              4                        = RT1ln2 = 2149 J.
                V0              V0 V
              2               4
The total work of gases compressing is:
        A = A12+A23+A34 = 9545 J ≅ 9.55 kJ              (1)
    c. In the second stage 23, all the water vapor (one mole) condenses. The heat Q’ delivered
in the process equals the sum of the work A and the decrease ∆U of internal energy of one
mole of water vapor in the condensing process.
                       Q' = ∆U + A12+A23+A34
    One can remark that ∆U + A23 is the heat delivered when one mole of water vapor
condenses, and equals 0.018×L. Thus

                                                                                1

 Q’ = ∆U + A11+A23+A34 = 0.018× L + A11 + A34 = 46.946 J ≅ 47 kJ         (2)

2. The process of compression (2. a.) and expansion (2.c.) of gases can be divided into several
stages. The stages are limited by the following states:
 State    Left compartment          Right compartment         Total        Pressure on
         Volume | Pressure         Volume | Pressure         volume        piston (atm.)
                 | (atm.)                   | (atm.)
   1     V0      | 0.5             V0      | 0.5              2V0              0.5
   2     V0      | 0.5             0.5 V0 | 1                 1.5V0            1
   3     0.5V0 | 1                 V0/3     | 1.5            5/6 V0            1.5
   4     0       | 1               V0/3     | 1.5             V0/3             1.5
   5     0       | 1.5             V0/4     | 2               V0/4              2
   6     0       | 1.5             V0/3     | 1.5             V0/3             1.5
   7     0       | 1               V0      | 0.5               V0              0.5
   8     0.5V0 | 1                 V0      | 0.5             1.5V0             0.5
   9     V0(2- 2 )|( 2 +2)/4       V0 2 |        2/4         2V0                2 / 4 = 0.35

a. See figure 2 below.

b. The work Ap done by the piston in the process of compressing the gases equals the sum of
the work A calculated in 1. and the work done by the force of friction. The latter equals
0.5 atm.×V0 = p1V0 (the force of kinetic friction appears in the process 234 during which the
displacement of the partition NM corresponding to a variation V0 of volume of the left
compartment). Then, we have:
         Ap = A + p1V0 = 9545 + 8.31 ×373 = 12645 J ≅ 12.65 kJ

c. In process 89 the pressure in the left compartment is always larger than the pressure in the
right one (with a difference of 0.5 atm.). If p denotes the pressure in the right compartment,
the pressure in the left one will be p + 0.5 atm.
    Let V be the total volume in process 89, we have
                 RT1      RT1
                      +          =V
                  p      p + 0.5
                2 RT1                          RT1    RT1      2 RT1
with V = 2V0 =         , p can be defined by       +         =
                 0 .5                           p    p + 0.5    0.5
This is equivalent to:
                p + 0.5 + p = 4p.(p+0.5)
                       1       2
                 p=       =      = 0.35 atm.
                        8     4
The pressure in the right compartment is p = 0.35 atm.
The volume of the right compartment is 2 V0
The pressure in the left compartment is p + 0.35 = 0.85 atm.
The volume of the left compartment is (2 - 2 ) V0

                                                                                2

    p(atm)

                 5
         2

                         4        3
       1.5
                     6
                                                        2
        1

       0.5                                                            1
      0.35                               7              8
                                                                      9

                 V0 V0       V0   5V0
             0                          V0              1.5V0       2V0      V
                  4 3         2    6

                                             Figure 2

3.
a. Apply the first law of thermodynamic for the system of two gases in the cylinder:
                     δq = dU + δA                                    (3)
    In an element of process in which the variations of temperature and volume are
respectively dT and dV:
                                               R       R
    δq = 0 ; dU = (CV1 + CV2)dT = (                +       )dT ; δA = p.dV
                                             γ1 −1 γ 2 −1
On the other hand:                    pV = RT                         (4)
One can deduce the differential equation for the process:
                              R        R
                          (       +         )dT + p. dV = 0
                            γ1 −1 γ 2 −1
or
                            dT (γ 1 − 1)(γ 2 − 2) dV
                                +                     =0                (5)
                            T       γ1 + γ 2 − 2 V
By putting
                               K= 1
                                    (γ − 1)(γ 2 − 1) = 2               (6)
                                      γ1 + γ 2 − 2     11
after integrating (5), we have:
                                    TVK = const.                       (7)
The condensing temperature of water vapor under the pressure 0.5 atm. is also the boiling
temperature T’ of water under the same pressure. By using the given approximate formula, we
obtain
                1 1            R      p
                   − = (− ) ln
                T ' T0        µL     p0

                                                                                3

•      If we consider p approximatively constant (with relative deviation about 20/373 ≈ 5 %)
T ' can be easily found, and
                         T’ = 354 K
The volume V ' of the right compartment at temperature T ' can be calculated as follows:
                               1             11
                              T K          373  2
                   V ' = V0  1  = V0            = 1.33 V0 ≅ 1.3 V0
                             T'            354 
                                       = 81.6 dm3 = 0.0816 m3 ≅ 0.08 m3
b.The work done by the gas in the expansion is:
                                                  R         R
         A = - ∆U = (CV1+CV2)(T0- T ' ) = (            +        )(373 − 354) =
                                               γ 1 −1 γ 2 − 2
                                               5      6
                                           = ( R + R )×19 = 868 J ≅ 9×102 J
                                               2      2
• If we consider the dependence of water vapor pressure p on temperature T ' , we must
resolve the transcendental equation
                     1 1             R     1 T'       R        R T'
                        −     = (− ) ln           =     ln 2 −     ln
                    T ' T0          µL 2 T0 µL                 µL T0
This equation can be reduced to a numerical one:
                    1     1                                          T'
                        −       = 1.422 × 10−4 − 2.052 × 10−4 ln
                    T ' 373                                         373
By giving T ' different values 354, 353, 352 and choosing the one which satisfies this
equation, we find the approximate solution :
                             T ' = 353 K
With this value of temperature, the volume of the right compartment is :
                              11
                        373 2
             V ' = V0(      ) = 1.35 V0 = 0.082 m3 ≅ 0.08 m3
                        353
b. The work done by the gas in the expansion is:
                 11
            A=      R×20 = 914 J ≅ 9×102 J
                  2

                                                                                4

APhO 2004 Theory T3


[User]
Newton’s rings are an interference pattern formed when a lens is placed on a flat
    glass surface and illuminated from above by light of wavelength λ. For concreteness, suppose the
    side of the lens touching the surface is spherical, with radius of curvature R. When viewed from
    above, one sees an interference pattern with circular fringes.

    The most important reflection paths are (1) reflection off the top surface of the lens, (2) reflection
    off the bottom surface of the lens, and (3) reflection off the flat surface. However, the first reflection
    path has a very different path length from the other two, which means it won’t give rise to visible
    interference fringes, as explained in the remark above. Instead, the first reflection path just adds
    some background intensity everywhere, preventing the dark fringes from being perfectly dark. Thus,
    in this problem we’ll only consider the second and third paths.

      (a) Explain why the center of the pattern is dark.

      (b) Find the radii of the bright and dark fringes, i.e. the values of r where there is a local minimum
          or maximum of the intensity. For simplicity, assume r ≪ R.

[3] Problem 9 (Kalda). A hall of a contemporary art installation has white walls and a white ceiling,
    lit with a monochromatic green light of wavelength λ = 550 nm. The floor of the hall is made of
    flat transparent glass plates. The lower surfaces of the glass plates are matte and painted black; the
    upper surfaces are polished and covered with thin transparent film. A visitor standing in the room
    will see circular concentric bright and dark strips on the floor, centered around himself. A curious
    visitor observes that the stripe pattern depends on their height, and upon lowering themselves, sees
    a maximum of 20 stripes. The film’s index of refraction is 1.4 and the glass’s is 1.6. Determine the
    thickness of the film.

    3    Diffraction
    Next, we turn to diffraction.

        Example 4

        Find the interference pattern of a diffraction grating, a set of N identical slits in a row, each
        separated by a distance d.

        Solution
        Defining ∆r = d sin θ as before, the amplitude is

                                                                           eikN ∆r − 1
                          A ∼ 1 + eik∆r + e2ik∆r + . . . + e(N −1)ik∆r =               .
                                                                            eik∆r − 1
        Factoring out a common phase, we have

                                       eikN ∆r/2 − e−ikN ∆r/2   sin(N k∆r/2)
                                 A∼       ik∆r/2    −ik∆r/2
                                                              =
                                         e       −e              sin(k∆r/2)

       so the intensity is
                                                 sin2 (N k∆r/2)
                                            I∝                  .
                                                  sin2 (k∆r/2)
       The normalized intensity is plotted below as a function of θ.

       The numerator yields rapid oscillations which aren’t very visible; their envelope is given by
       the slow oscillations in the denominator. These slow oscillations are the ones we care about;
       they are the diffraction peaks and occur when
                                k∆r                       2πn
                                    = nπ,     d sin θ =       = nλ,    n ∈ Z.
                                 2                         k
       This is intuitive, because at the maxima, the contributions from each slit are in phase, as
       the path length difference between adjacent slits is a multiple of λ.

Solution:
(a) Consider light that comes in very close to the center. Then paths (2) and (3) have
         almost the same path length, but path (2) has reflection from glass to air, which comes with
         no phase shift, and path (3) has reflection from air to glass, which comes with a π phase shift.
         Therefore, the paths destructively interfere, and the center is dark.

      (b) Since r ≪ R, we can approximate the light as going straight up and down. Putting the origin
          at the place the lens and flat surface touch, the equation of the lens’s curved surface is

                                                 r2 + (y − R)2 = R2

          where r is the distance from the axis of symmetry. We thus have

                                                   r2 = 2yR − y 2 .

          We know that r ≪ R, so for the left-hand side to match the right-hand side, we must have
          y ≪ R, which in turn implies the y 2 term is negligible. Dropping it gives

                                                            r2
                                                       y≈      .
                                                            2R
          The path length difference is 2y, and we have an extra π phase shift as explained in part (a),
          so the condition for destructive interference is
                                                r2                 √
                                                   = nλ,    r=         nλR
                                                R
          while the condition for constructive interference is
                                       r2                          p
                                          = (n + 1/2)λ,     r=      (n + 1/2)λR.
                                       R
          This is a practical way to quickly check how spherical a lens really is.

[3] Problem 9 (Kalda). A hall of a contemporary art installation has white walls and a white ceiling,
    lit with a monochromatic green light of wavelength λ = 550 nm. The floor of the hall is made of
    flat transparent glass plates. The lower surfaces of the glass plates are matte and painted black; the
    upper surfaces are polished and covered with thin transparent film. A visitor standing in the room
    will see circular concentric bright and dark strips on the floor, centered around himself. A curious
    visitor observes that the stripe pattern depends on their height, and upon lowering themselves, sees
    a maximum of 20 stripes. The film’s index of refraction is 1.4 and the glass’s is 1.6. Determine the
    thickness of the film.

Solution. Both rays will bounce off a hard surface, so the phase shift due to that will be ignored.
All that will be considered is the difference in optical path lengths.

   Note that sin α = n sin β. For the immediately reflected ray, it will travel through L1 = w sin α,
and tan β = (w/2)/t, so L1 = 2t tan β sin α. For the ray that goes through the film, it will travel by
L2 = 2nt/ cos β where n = 1.4, so the path length difference is

              2nt    2t sin β sin α        1 − sin2 β      p                    p
      ∆L =         −                = 2nt p            = 2t n2 − (n sin β)2 = 2t n2 − sin2 α.
             cos β       cos β              1 − sin2 β
                               √
So ∆L ranges from 2tn to 2t n2 − 1. Constructive interference is where ∆L       √ = λk with k as an
         Thus for this situation, there are 20 values of λk that fits between 2t n2 − 1 and 2tn, thus
integer. √
2tn − 2t n2 − 1 ≈ 20 λ, giving
                                              20 λ
                                    t≈        √        ≈ 13 µm.
                                         2(n − n2 − 1)

3    Diffraction
Next, we turn to diffraction.

    Example 4

    Find the interference pattern of a diffraction grating, a set of N identical slits in a row, each
    separated by a distance d.

    Solution
    Defining ∆r = d sin θ as before, the amplitude is

                                                                       eikN ∆r − 1
                      A ∼ 1 + eik∆r + e2ik∆r + . . . + e(N −1)ik∆r =               .
                                                                        eik∆r − 1
    Factoring out a common phase, we have

                                  eikN ∆r/2 − e−ikN ∆r/2   sin(N k∆r/2)
                             A∼      ik∆r/2    −ik∆r/2
                                                         =
                                    e       −e              sin(k∆r/2)

       so the intensity is
                                                 sin2 (N k∆r/2)
                                            I∝                  .
                                                  sin2 (k∆r/2)
       The normalized intensity is plotted below as a function of θ.

       The numerator yields rapid oscillations which aren’t very visible; their envelope is given by
       the slow oscillations in the denominator. These slow oscillations are the ones we care about;
       they are the diffraction peaks and occur when
                                k∆r                       2πn
                                    = nπ,     d sin θ =       = nλ,    n ∈ Z.
                                 2                         k
       This is intuitive, because at the maxima, the contributions from each slit are in phase, as
       the path length difference between adjacent slits is a multiple of λ.


[User]
01^      USAPhO 2019, problem B3. It’s worth reading the solution carefully afterward.

    4    Rotational Collisions
        Idea 7: Angular Impulse

        During a collision with impulse J, the angular momentum changes by the “angular impulse”
        r × J. In many problems involving collisions which conserve angular momentum, energy is
        necessarily lost in the collision process. This is another example of an inherently inelastic
        process, an idea we first encountered in M3.

[3] Problem 19 (Morin 8.22). A uniform ball of radius R and mass m rolls without slipping with
    speed v0 . It encounters a step of height h and rolls up over it.

     (a) Assuming that the ball sticks to the step during this process, show that for the ball to climb
         over the step,
                                                             5h −1
                                               r
                                                  10gh
                                          v0 ≥          1−           .
                                                    7        7R

     (b) Energy is lost to heat by the inelastic collision of the ball with the step. In the limit of small
         h, how much heat is produced?

[3] Problem 20 (KK 6.38). A rigid massless rod of length L joins two particles, each of mass m. The
    rod lies on a frictionless table, and is struck by a particle of mass m and velocity v0 as shown.

   After an elastic collision, the projectile moves straight back. Find the angular velocity of the rod
   about its center of mass after the collision.

[3] Problem 21 (PPP 47). Two identical dumbbells move towards each other on a frictionless table
    as shown.

    Each consists of two point masses m joined by a massless rod of length 2ℓ. The dumbbells collide

                       01m
    elastically; describe what happens afterward.

Solution:
(a) Let β = 2/5. Once the ball collides with the corner, it essentially rotates around
        that corner, and we will first aim to find the initial angular velocity of the rotation of the
        ball around the corner. Note that angular momentum about the corner is conserved, since
        the only relevant force during the very short collision time is the large force applied at the
        corner, so the net torque is 0. This is an inherently inelastic process; energy is lost during
        this collision.
         Let us compute the angular momentum right before the collision occurs. It is just the sum of
         the angular momentum of the CM, plus the angular momentum around the CM, so
                                                        v0
                                            Li = βmR2      + Rmv0 (1 − h/R),
                                                        R
         since the sine of the angle between p and R is 1 − h/R. Let the angular velocity about the
         corner be ω. Then, the final angular momentum is

                                                  Lf = (1 + β)mR2 ω,

         so equating the two tells us that

                                        β + 1 − h/R                    1 h
                                   Rω =             v0 =         1−                  v0 .
                                           β+1                        β+1R

         Now, as the ball rotates about the corner, energy is conserved, so the only way that the ball
         will make it to the top is if its kinetic energy is at least mgh. Therefore,

                                                             1 h 2 2

                         1          2 2         1
                           (β + 1)mR ω ≥ mgh =⇒ (β + 1) 1 −        v0 ≥ gh,
                         2                      2           β+1r
         or                                     s                     −1
                                                    2gh          1 h
                                         v0 ≥              1−                ,
                                                    β+1         β+1r
         which is exactly what we wanted to show.

     (b) The initial kinetic energy of the ball is 12 (1 + β)mv02 . We can use the previously found equation

                                                                   1 h
                                           vf = Rω = 1 −                  v0 ,
                                                               β+1R

         which helps us find the kinetic energy immediately after the inelastic collision 12 (1 + β)mvf2 .
         Thus the kinetic energy dissipated into heat is
                                                                                !
                                                                              1 h 2

                            1            2    2    1          2
                     ∆Q = (1 + β)m(v0 − vf ) = (1 + β)mv0 1 − 1 −                         .
                            2                      2                        1+βR

         Using the binomial approximation, we conclude

                                                                   mv02 h

                                       1                 2h
                                 ∆Q ≈ (1 + β)mv02                =        .
                                       2              (1 + β)R      R

         Interestingly, the ratio of this to the amount of gravitational potential energy needed to climb
         the step, which is mgh, is independent of h. So even if we turn a big step into many tiny
         steps, it’ll still be substantially less efficient than a smooth slope.

[3] Problem 20 (KK 6.38). A rigid massless rod of length L joins two particles, each of mass m. The
    rod lies on a frictionless table, and is struck by a particle of mass m and velocity v0 as shown.

    After an elastic collision, the projectile moves straight back. Find the angular velocity of the rod
    about its center of mass after the collision.

    Solution. Suppose the projectile moves back with speed v1 , the CM speed of the dumbbell is v2 ,
    and its angular velocity about its CM is ω. Then, momentum, angular momentum, and energy
    conservation yield

                                 mv0 = −mv1 + 2mv2 =⇒ v0 + v1 = 2v2
                           √                      √              √
                   mv0 L/2 2 = (mL2 /2)ω − mv1 L/2 2 =⇒ v0 + v1 = 2Lω
                        mv02 = mv12 + 2mv22 + (mL2 /2)ω 2 =⇒ (v0 − v1 )(v0 + v1 ) = 3v22 .

    Combining the first and last equations implies v0 − v1 = (3/2)v2 , so 2v0 = (7/2)v2 , so v2 = 47 v0 .
    Using the second equation gives
                                                 √             √
                                        √       4 2           4 2 v0
                                  Lω = 2v2 =        v0 , ω =          .
                                                 7              7 L
[3] Problem 21 (PPP 47). Two identical dumbbells move towards each other on a frictionless table
    as shown.

    Each consists of two point masses m joined by a massless rod of length 2ℓ. The dumbbells collide
    elastically; describe what happens afterward.

    Solution. First we find out what happens immediately after the collision. Suppose the dumbbells
    move in the opposite direction at v1 , and have angular velocity ω > 0.

   Angular momentum conservation tells us 4mℓ2 ω − 4mv1 ℓ = 4mvℓ, so v1 + v = ℓω. Energy conserva-
   tion tells us 2mv12 + 2mℓ2 ω 2 = 2mv 2 , so (v − v1 )(v + v1 ) = ℓ2 ω 2 . Therefore, v − v1 = ℓω, so v1 = 0.
   Thus, the rods both rotate at angular velocity v/ℓ.
      Once both rods rotate 180◦ , they collide again. By using the reasoning of the first collision in
   reverse, the rods simply lose their angular velocity and regain their original translational velocities.
   Therefore, the final result is that both rods translate uniformly, as if they passed right through each
   other, but both rods are flipped upside down.


[User]
Kondensaatorite süsteem koosneb viiest kondensaatorist. Alghetkel on kolm neist laenguta ning kahel paikneb laeng $q$ (vt joonist). Missugune laeng koguneb keskmisele kondensaatorile, kui süsteem on jõudnud tasakaaluolekusse?

\begin{center}
	\includegraphics[width=0.42\linewidth]{2009-v3g-03-G_kondensaatorid.pdf}
\end{center}

Solution:
Kogunegu keskmisele kondensaatorile (mahtuvusega $3C$) laeng $a$ ning nurgas paiknevatele kondensaatoritele (mahtuvusega $3C$) laeng $b$. Vaatleme ülemist vasakpoolset kondensaatorit: selle negatiivsel plaadil on nüüd laeng $-q+a+b$ ning positiivsel plaadil $q+b$. Saame võrrandi:
\[-(-q+a+b)=q+b \implies q-a-b=q+b \implies b=-a/2.\]
Lisaks saame pingete võrdsusest
\[\frac{a}{3C}=\frac{q+b}{C}+\frac{b}{3C}\implies a=3q+4b\implies a+2a=3q,\]
millest $a=q$.

Kondensaatorid

Kondensaatorite süsteem koosneb viiest kondensaatorist. Alghetkel on kolm neist laenguta ning kahel paikneb laeng $q$ (vt joonist). Missugune laeng koguneb keskmisele kondensaatorile, kui süsteem on jõudnud tasakaaluolekusse?

\begin{center}
	\includegraphics[width=0.42\linewidth]{2009-v3g-03-G_kondensaatorid.pdf}
\end{center}

Süsteemis kehtib laengu jäävus, st alamsüsteemis, mis koosneb ülemisest sõlmpunktist ja sellega ühendatud kolmest kondensaatori plaadist peab alati olema summaarse laenguga $-q$. See kehtib sellepärast, et laengud saavad liikuda ainult mööda metalli ning ei saa eelmainitud alamsüsteemist õhu kaudu lahkuda.


[User]
The motion of ripples of short wavelength (less than 1 cm) on water is controlled by
    the surface tension γ and density ρ.

      (a) Find the phase velocity vp of ripples with wavenumber k, up to a dimensionless constant.

      (b) Show that vg = (3/2)vp .

    3      Reflection and Transmission
    When we considered standing waves in the previous section, we were only considering “steady state”
    behavior. Now we consider the dynamics of a wave hitting an obstacle more explicitly.

         Example 3

         Suppose a string defined for x < 0 ends at a hard wall at x = 0. Show that any wave directed
         towards the wall will be reflected back upside-down.

         Solution
         We suppose that we send in a wave of the form

                                           yin (x, t) = f (kx − ωt).

         Let the reflected wave be a general wave traveling backward,

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

       Both of these expressions only have physical meaning for x < 0, since the string only exists
       there. Now, the boundary condition is y(0, t) = 0, so we have

                                           f (−ωt) + g(−ωt) = 0.

       This tells us precisely that g = −f , so the wave is reflected upside-down but otherwise
       unchanged.

       There’s an easy way to visualize what’s going on here. We can imagine that there really is
       string for x > 0, but that the point x = 0 stays fixed for some reason. Then this situation
       corresponds to an incoming wave coming from the left, and a flipped wave coming from the
       right. The two meet and cancel at x = 0, and the flipped wave continues on going to the
       left, where the physical string is. Fundamentally, this story works for the same reason as the
       method of images in electromagnetism: as long as you satisfy the boundary conditions, you
       can do whatever you want beyond the boundary.

Solution:
(a) By dimensional analysis, the only possible dispersion relation is
                                                          s
                                                            γk 3
                                                  ω(k) ∝         .
                                                             ρ

          This tells us that                                  s
                                                      ω           γk
                                                  vp = ∝             .
                                                      k            ρ

     (b) We have                                                s
                                                      dω   3        γk
                                                 vg =    ∝
                                                      dk   2         ρ
          with the same constant of proportionality as in part (a), giving the desired result. So inter-
          estingly, this is a case where dimensional analysis can give us a numeric prefactor!

    3    Reflection and Transmission
    When we considered standing waves in the previous section, we were only considering “steady state”
    behavior. Now we consider the dynamics of a wave hitting an obstacle more explicitly.

        Example 3

        Suppose a string defined for x < 0 ends at a hard wall at x = 0. Show that any wave directed
        towards the wall will be reflected back upside-down.

        Solution
        We suppose that we send in a wave of the form

                                            yin (x, t) = f (kx − ωt).

        Let the reflected wave be a general wave traveling backward,

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

       Both of these expressions only have physical meaning for x < 0, since the string only exists
       there. Now, the boundary condition is y(0, t) = 0, so we have

                                           f (−ωt) + g(−ωt) = 0.

       This tells us precisely that g = −f , so the wave is reflected upside-down but otherwise
       unchanged.

       There’s an easy way to visualize what’s going on here. We can imagine that there really is
       string for x > 0, but that the point x = 0 stays fixed for some reason. Then this situation
       corresponds to an incoming wave coming from the left, and a flipped wave coming from the
       right. The two meet and cancel at x = 0, and the flipped wave continues on going to the
       left, where the physical string is. Fundamentally, this story works for the same reason as the
       method of images in electromagnetism: as long as you satisfy the boundary conditions, you
       can do whatever you want beyond the boundary.


[User]
Solenoid
A solenoid of length l = 20 cm is wound around a vertical, cylindrical test tube made of glass,
filled with water. The solenoid is thermally insulated from the water. The height of the water
level is approximately 20 cm above the upper end of the solenoid, the diameter of the test tube is
1 cm, the number of turns of the coil is N = 6000.

   The atmospheric pressure is p0 = 101 kP a, the temperature of water is 293 K. Magnetic
susceptibility of water is χ ≡ μr − 1 = −9.04 · 10−6 . Vacuum permeability μ0 = 12.57 · 10−7 H/m.
The current in the solenoid is slowly increased until the water starts boiling inside the coil. At
which current strength would this happen? Make reasonable approximations when needed. Note
that the required current strength might be slightly too large for the present technology.

Solution:
Solution - Solenoid
    Magnetic field gives rise to magnetization of water molecules, i.e. each of the molecules becomes
a magnetic dipole. Inhomogeneous magnetic field exerts a force to a magnetic dipole. Water
molecules being diamagnetic are pushed away from the region of stronger magnetic field. Water
reaches a state with mechanical equilibrium where the magnetic force is balanced by a force
due to pressure gradient. Hence, in mechanical equilibrium, regions with stronger magnetic field
correspond to smaller pressures. When the current in the solenoid is increased, the pressure
differences grow, and at a certain moment, a region will appear where the pressure is smaller than
the saturation pressure of the water vapour. This is the moment when water starts boiling.

   To begin with, let us discuss possible reasonable approximations. First, we can neglect the water
column pressure which is only 2 to 4 percents of the atmospheric pressure. Second, the saturation
pressure of water vapour under normal conditions is also much smaller than the atmospheric one
and therefore can be neglected. Thus we can say that boiling starts when the pressure drop due
to magnetic field becomes equal to p0 .

    So, we need to relate the pressure difference caused by the magnetic field to the magnetic field
strength. Notice that if a region with magnetic field B is filled with water, the magnetic field
energy density is B 2 /(2μr μ0 ); meanwhile, if there is no water, the energy density is B 2 /(2μ0 ). So,
we can ascribe the energy density difference

                                                 B2

                                                        1
                                        ∆w =               −1
                                                 2μ0 μr

to the interaction of water and magnetic field. Next, consider the following thought experiment.
We push away a small volume V of water from the neighbourhood of a point P in the water
where the magnetic field strength is B; the displaced water fills in a narrow layer at the top of the
water surface of equal volume. Assuming that the magnetic field is negligibly small at the top,
by comparing the initial and final states, we conclude that the total interaction energy is reduced
by ∆wV . When pushing away water from P we perform mechanical work pV , where p is the
pressure at point P . At the upper surface of the water, the moving interface performs mechanical
work p0 V so that the net mechanical work performed by water during this process is (p0 − p)V .
Due to energy conservation law, ∆wV = (p0 − p)V so that:

                                               B2

                                                     1
                                     p0 = p +           −1 .
                                               2μ0 μr

Note that the way how we derived this relationship is completely analogous to how the Bernoulli
law is derived, and in fact, the obtained equality can be interpreted as a modified Bernoulli law
for zero speed where the volume density of potential energy in gravity field ρgh is replaced with
∆w - the energy density of magnetic interaction. This equality can be simplified by noting that
 1
    − 1 = −χ/μr ≈ −χ so that:
μr
                                                    χB 2
                                          p0 = p −       .
                                                    2μ0

As discussed above, the boiling condition is p ≈ 0, hence:
                                               r
                                                 −2μ0 p0
                                         B=              .
                                                    χ
                                                                                N
Finally, we apply the formula for magnetic field strength inside a long solenoid B = μ0 I to find:
                                                                                `
                                           r
                                         `      2p0
                                    I=               = 4.4 kA.
                                        N −χμ0

Solenoid


[User] Find the mass of the water that has to be poured into a cube so that the cube would be as stable as possible, meaning that the system’s center of mass would be as low as possible. The length of the cube’s edge is $a$, the mass $M$, density of water $\rho$. Do not consider the width of the cube’s wall. The cube is completely symmetrical meaning that it has all 6 identical faces.

Solution:
\emph{Esimene lahendus}\\
Tähistame otsitava vee massi $m$, veesamba kõrguse $h$ ning süsteemi massikeskme kõrguse $l$. Kuna kuup on sümmeetriline, asub selle massikese kõrgusel $\frac{a}{2}$. Vee massikese asub veekoguse keskel ehk kõrgusel $\frac{h}{2}$. Kui kuup on tühi, siis ühtib süsteemi massikese kuubi massikeskmega ehk $l=\frac{a}{2}$ ning veesamba kõrgus $h=0$. Kui nüüd kuubi põhja aeglaselt vett valada, hakkab veesamba kõrgus $h$ kasvama ning süsteemi massikeskme kõrgus $l$ vähenema, kuna kogu lisatud vesi asub algsest süsteemi massikeskmest all pool. Süsteemi massikeskme kõrgus $l$ ei saa vee lisamisega enam alaneda, kui see ühtib veesamba kõrgusega $h$, sest kui selles olukorras vett juurde lisada, oleks äsja juurde lisatud veekogus eelnevast süsteemi massikeskmest kõrgemal ning süsteemi massikeskme kõrgus hakkaks kasvama. Seega, massikese on võimalikult madalal olukorras, kui süsteemi massikeskme kõrgus ühtib veesamba kõrgusega ehk $l=h$.
Rakendades kangi reeglit saame:
$$M(\frac{a}{2}-l)=m(l-\frac{h}{2}).$$
Teades seost $l=h$ ja vee massi $m=\rho a^2h$ ning asendades need eelnevasse võrrandisse:
$$M(\frac{a}{2}-h)=\rho a^2h(h-\frac{h}{2}).$$
Siit saame $h$ jaoks ruutvõrrandi:
$$\rho a^2h^2+2Mh-Ma=0,$$
$$h=\frac{-2M \pm \sqrt{4M^2+4\rho a^3M}}{2\rho a^2}.$$
Kuna negatiivne lahend ei sobi, saame
$$h=\frac{M(\sqrt{1+\frac{\rho a^3}{M}}-1)}{\rho a^2}.$$
Otsitav veekoguse mass on seega
$$m=\rho a^2h=M(\sqrt{1+\frac{\rho a^3}{M}}-1).$$

\emph{Teine lahendus}\\
Minimaalsele süsteemi massikeskmele vastavat veesamba kõrgust on võimalik leida ka tuletise abil. Kuubi sümmeetria tõttu asub selle massikese kõrgusel $\frac{a}{2}$ ning vee massikese kõrgusel $\frac{h}{2}$. Rakendades kangi reeglit saame:
$$M(\frac{a}{2}-l)=m(l-\frac{h}{2}).$$
Sellest tuleb avaldada süsteemi massikeskme kõrgus $l$ ning otsida $h$-d, kui $\frac{dl}{dh}=0$. Teades, et vee mass on $m=\rho a^2h$:
$$l=\frac{Ma+\rho a^2h^2}{2(M+\rho a^2h)},$$
$$\frac{dl}{dh}=\frac{4\rho a^2h(M+\rho a^2h)-2\rho a^2(Ma+\rho a^2h^2)}{4(M+\rho a^2h)^2}=0.$$
Siit saame lihtsustades ning $\rho a^2$-ga läbi jagades $h$ jaoks ruutvõrrandi:
$$\rho a^2h^2+2Mh-Ma=0.$$
See on identne eelnevalt saadud ruutvõrrandiga ning seega on ka saadav vastus on sama:
$$m=M(\sqrt{1+\frac{\rho a^3}{M}}-1).$$

Kuup veega

Leidke veekoguse mass, mis tuleb valada kuupi, et see oleks võimalikult stabiilne, ehk süsteemi massikese oleks võimalikult madalal. Kuubi külje pikkus on $a$, mass $M$, vee tihedus $\rho$. Kuubi seina paksusega mitte arvestada. Kuup on täielikult sümmeetriline ehk sellel on olemas kõik 6 identset tahku.

Süsteemi massikese on võimalikult madalal siis, kui veesamba kõrgus ühtib süsteemi massikeskme kõrgusega. Selles saab veenduda, kui vaadelda, mis juhtub väikese veekoguse lisamisel või eemaldamisel.