[User]
Some nuclei have extremely long lifetimes τ , so that we can measure τ by continuously
    watching a very large sample of N0 nuclei, and looking for decay events. However, it turns out that
    the way we do it can yield different results. Let’s consider the following procedures.

      (a) We start a stopwatch at noon and stop it when the next decay happens, giving t1 .

     (b) We have an intern watch the sample continuously, then at noon, ask them how long it was
         since the last decay, giving t2 .

      (c) We have an intern watch the sample continuously, then at noon, ask them how long it was
          since the last decay. We then set our stopwatch so that t = 0 when that decay happened, and
          stop the stopwatch when the next decay happens, giving t3 .

     (d) We continuously watch the sample, start a stopwatch when the first decay happens, then stop
         it when the next decay happens, giving t4 .

    We repeat procedure i many times, so the average of ti is ti . Find the ti in terms of N0 and τ .

[2] Problem 5 (Krane 12.37). A radioactive sample contains N0 atoms at time t = 0. It is observed
    that N1 radioactive atoms remain at time t1 and then decay by time t2 , N2 remain at t2 and then
    decay by time t3 , and so on. Show that if many observations are made, then τ can be measured as
                                                    1 X
                                               τ=       Ni ti .
                                                    N0
                                                            i

        Example 1

        Radium can be found in trace quantities throughout the Earth, and has a half-life of 1620
        years. Suppose that there is currently 1 kg of radium on the Earth. Then extrapolating

        backwards, there was 24.5×10 /1620 kg of radium on the Earth when it was formed, which is
        greater than the mass of the observable universe! What’s wrong with this calculation?

        Solution
        Nuclear decays don’t happen in isolation; there are entire networks of nuclear decay chains.
        Radium decays quickly, but it is also constantly produced by the decay of other isotopes,
        which have much longer half-lives.

Solution:
(a) The probability of any decay in a time interval dt is N0 dt/τ , so the probability
         of having no decay in that interval is (1 − N0 dt/τ ). After N = t/dt such time intervals, the
         probability that a single decay still hasn’t occurred is (1 − N0 dt/τ )N . As shown in P1, this
         becomes e−N0 t/τ in the limit dt → 0. Thus, the probability of the first decay occurring after
         time t in an interval dt is
                                                         N0 −N0 t/τ
                                             P (t) dt =     e       dt.
                                                          τ
         The value of t1 is the average of this time, so
                                     N0 ∞ −N0 t/τ
                                         Z                    Z ∞
                                                                                τ
                                t1 =          e        t dt =     e−N0 t/τ dt =    .
                                      τ 0                      0                N0

     (b) “Waiting” forward or backwards in time are symmetric, so t2 = t1 = τ /N0 . (Technically,
          there’s a tiny difference because the previous decay occured when there were N0 + 1 nuclei
          instead, but this is negligible in a typical sample containing billions of billions of nuclei.)

      (c) By definition, t3 = t1 + t2 , and taking expectation values gives t3 = t1 + t2 . Thus, t3 = 2τ /N0 .

     (d) We know that the mean time between decays is τ /N0 , so t4 = τ /N0 .
          Of course, the tricky part of the problem is the following: why is t3 ̸= t4 , even though they
          seem to be measuring the exact same thing, namely the time between two decays? The
          difference is in the way we select the decay we look at. For t4 , we look at a random decay
          event (i.e. if there are a thousand decay events, each one has an equal chance of being the one
          we look at). But for t3 , we look at the decay happening during a random time, which means
          that longer time intervals have a larger chance of being randomly picked, so t3 > t4 .
          To show this explicitly, note that the probability distribution of decay times is (N0 /τ )e−N0 t/τ ,
          as derived in part (a). The probability distribution of decay times weighted by decay length,
          as used in part (c), is (N0 /τ )2 t e−N0 t/τ . So the expected decay time in part (c) is

                                                N02 ∞ 2 −N0 t/τ
                                                      Z
                                                                            τ
                                          t3 = 2           t e      dt = 2
                                                 τ      0                  N0

          just as argued more intuitively above.
         This is quite a tricky factor of 2. Drude got it wrong when formulating the Drude model,
         which is the simplest classical model of electrical conduction in a metal. It turns out that the
         Drude model is totally wrong, due to quantum mechanics, but this mistake, plus two other
         more conceptual issues, made it look like it agreed with experiment.
       Another example of a memoryless process is the collisions of a given gas molecule in an ideal gas,
    according to kinetic theory. For example, all of the subparts above could have been rephrased in
    terms of observing the distance a gas molecule moves between collisions, with the same conclusions.

[2] Problem 5 (Krane 12.37). A radioactive sample contains N0 atoms at time t = 0. It is observed
    that N1 radioactive atoms remain at time t1 and then decay by time t2 , N2 remain at t2 and then
    decay by time t3 , and so on. Show that if many observations are made, then τ can be measured as
                                                     1 X
                                                τ=       Ni ti .
                                                     N0
                                                            i

    Solution. N (t) should follow N (t) = N0 e−t/τ , so dN (t)/dt = −N (t)/τ . Thus the number of atoms
    that decay between time ti and ti+1 , Ni , will be about Ni = (ti+1 − ti )dN (ti )/dt = (ti+1 − ti )N (t)/τ
    as the number of measurements are large. With smaller time intervals, this can be seen as Ni =
    N (t)dt/τ . Thus looking at the expression N10 i Ni ti gives
                                                     P

                                                Z ∞
                                                                  1 ∞ −t/τ
                                                                  Z
                            1 X         1                 dt
                                Ni ti ≈             N (t)      t=     e    tdt.
                            N0          N0       0        τ       τ 0
                                  i

    This integral can be evaluated with parts (differentiating t and integrating e−t/τ dt),

                                    1 ∞ −t/τ
                                      Z               Z ∞
                                          e     tdt =      e−t/τ dt = τ,
                                    τ 0                 0

    which shows that, as desired,
                                                     1 X
                                                τ=       Ni ti .
                                                     N0
                                                            i

       Example 1

       Radium can be found in trace quantities throughout the Earth, and has a half-life of 1620
       years. Suppose that there is currently 1 kg of radium on the Earth. Then extrapolating

       backwards, there was 24.5×10 /1620 kg of radium on the Earth when it was formed, which is
       greater than the mass of the observable universe! What’s wrong with this calculation?

       Solution
       Nuclear decays don’t happen in isolation; there are entire networks of nuclear decay chains.
       Radium decays quickly, but it is also constantly produced by the decay of other isotopes,
       which have much longer half-lives.

                  01W
                     01^


[User]
IPhO 1997, problem 3. A simple, neat question about how a plane works.

         Remark
         As you saw in the above problem, a plane works by having a wing angled so that air passing
         next to it gets pushed downward; momentum conservation then implies there is an upward

       force on the wing. Concretely, this upward force occurs because there is a higher pressure
       on the bottom of the wing, which by Bernoulli’s principle means there must be a higher air
       velocity on the top of the wing.

       This is all standard knowledge among engineers. However, in outdated or poor physics
       textbooks, an incorrect explanation is sometimes given. The story is that because the top
       side of the wing is curved, the air at the top of the wing must be moving faster, so that it
       can “catch up” to the air at the bottom. Then by Bernoulli’s principle there is a higher
       pressure on the bottom of the wing, causing lift.

       The second step of this argument is right, but the first step doesn’t make sense: there
       is no reason air should want to “catch up” with the air it used to be next to. For
       example, airplanes can fly upside down as long as the wing is angled the right way. In this
       case, the air at the top of the wing still moves faster, but now it’s the bottom side that’s curved.

       There’s much more to say about flight, which is a whole field of study. For some nice further
       discussion, see section 4.6 of The Art of Insight, or this paper and this paper by the same
       author. And for a perspective from a cantankerous old aerodynamicist, see this talk.

Solution:
IPhO 1997, problem 3. A simple, neat question about how a plane works.

       Remark
       As you saw in the above problem, a plane works by having a wing angled so that air passing
       next to it gets pushed downward; momentum conservation then implies there is an upward
       force on the wing. Concretely, this upward force occurs because there is a higher pressure
       on the bottom of the wing, which by Bernoulli’s principle means there must be a higher air
       velocity on the top of the wing.

       This is all standard knowledge among engineers. However, in outdated or poor physics
       textbooks, an incorrect explanation is sometimes given. The story is that because the top
       side of the wing is curved, the air at the top of the wing must be moving faster, so that it
       can “catch up” to the air at the bottom. Then by Bernoulli’s principle there is a higher
       pressure on the bottom of the wing, causing lift.

       The second step of this argument is right, but the first step doesn’t make sense: there
       is no reason air should want to “catch up” with the air it used to be next to. For
       example, airplanes can fly upside down as long as the wing is angled the right way. In this
       case, the air at the top of the wing still moves faster, but now it’s the bottom side that’s curved.

       There’s much more to say about flight, which is a whole field of study. For some nice further
       discussion, see section 4.6 of The Art of Insight, or this paper and this paper by the same
       author. And for a perspective from a cantankerous old aerodynamicist, see this talk.

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


[User]
A light originating from an object first goes through a concave lens and then a convex lens. The locations of the object, the convex lens and the forming image are shown in the figure, the focal points $F_k$ of the convex lens are also included. Construct the locations of the concave lens and the focal point $F_n$ that is closer to the object.\\
\begin{resizebox}{\linewidth}{!}{
		\begin{tikzpicture}
		\pgfmathsetmacro{\fk}{2}
		\pgfmathsetmacro{\tk}{3*\fk}
		\pgfmathsetmacro{\nk}{1/(1/\fk-1/\tk)}
		\pgfmathsetmacro{\a}{2*\fk}
		\pgfmathsetmacro{\h}{0.5}
		\pgfmathsetmacro{\htk}{\tk/\nk*\h}
		\pgfmathsetmacro{\ha}{1.5*\h}
		\pgfmathsetmacro{\nl}{\nk-\h/(\h-\ha)*(\nk-\a)}
		\pgfmathsetmacro{\fn}{\nk-\h/(\h-\ha)*(\nk-\nl)}
		\coordinate (TK) at (\tk, -\htk);
		\coordinate (NK) at (-\nk, \h);
		\coordinate (A) at (-\a, \ha);
		\draw (-1.1*\fn, 0) -- (1.1*\tk, 0);
		\draw[<->, ultra thick] (0,3.5*\h) -- (0,-3.5*\h);
		\draw[->, ultra thick] (\tk, 0) -- (TK);
		\draw[fill] (-\fk, 0) circle (0.02*\fk) node[below] {$F_k$};
		\draw[fill] (\fk, 0) circle (0.02*\fk) node[below] {$F_k$};
		\draw[->, ultra thick] (-\a, 0) -- (A);
		\end{tikzpicture}}
\end{resizebox}

Solution:
Nõguslääts tekitab esemest näiva kujutise ning kumerlääts tekitab näivast kujutisest tõelise kujutise. Teades kumerläätse asukohta ja fookuseid ning tõelise kujutise asukohta, on võimalik leida näiva kujutise asukoht (joonisel kujutatud sinisega; kujutatud on kolme kiirt, kuid konstrueerimiseks piisab kahest). Teades eseme ning nõgusläätse tekitatud näiva kujutise asukohti, on läätse asukoha ning selle esemepoolse fookuse leidmine lihtne.

\begin{resizebox}{\linewidth}{!}{
		\begin{tikzpicture}
		\pgfmathsetmacro{\fk}{2}
		\pgfmathsetmacro{\tk}{3*\fk}
		\pgfmathsetmacro{\nk}{1/(1/\fk-1/\tk)}
		\pgfmathsetmacro{\a}{2*\fk}
		\pgfmathsetmacro{\h}{0.5}
		\pgfmathsetmacro{\htk}{\tk/\nk*\h}
		\pgfmathsetmacro{\ha}{1.5*\h}
		\pgfmathsetmacro{\nl}{\nk-\h/(\h-\ha)*(\nk-\a)}
		\pgfmathsetmacro{\fn}{\nk-\h/(\h-\ha)*(\nk-\nl)}
		\coordinate (TK) at (\tk, -\htk);
		\coordinate (NK) at (-\nk, \h);
		\coordinate (A) at (-\a, \ha);
		\draw (-1.1*\fn, 0) -- (1.1*\tk, 0);
		\draw[<->, ultra thick] (0,3.5*\h) -- (0,-3.5*\h);
		\draw[->, ultra thick] (\tk, 0) -- (TK);
		\draw[->, ultra thick, blue] (-\nk, 0) -- (NK);
		\draw[blue] (TK) -- (NK);
		\draw[blue] (TK) -- (0,-\htk);
		\draw[blue] (0,-\htk) -- (NK);
		\draw[blue] (TK) -- (0,\h) -- (NK);
		\draw[fill] (-\fk, 0) circle (0.02*\fk) node[below] {$F_k$};
		\draw[fill] (\fk, 0) circle (0.02*\fk) node[below] {$F_k$};
		\draw[->, ultra thick] (-\a, 0) -- (A);
		\draw[>-<, ultra thick] (-\nl, 3.5*\h) -- (-\nl, -3.5*\h);
		\draw (-\nl, 0) -- (A) -- (NK);
		\draw (A) -- (-\nl, \ha);
		\draw[dashed] (-\nl, \ha) -- (-\fn,0);
		\draw[fill] (-\fn, 0) circle (0.02*\fk) node[below] {$F_n$};
		\end{tikzpicture}}
\end{resizebox}

Puuduv lääts

Esemelt lähtuv valgus läbib esmalt nõgusläätse ning seejärel kumerläätse. Joonisel on kujutatud eseme, kumerläätse ning lõpuks tekkiva kujutise asukohad ning kumerläätse fookused $F_k$. Konstrueerige lisalehel nõgusläätse ning selle esemepoolse fookuse $F_n$ asukohad.

\begin{resizebox}{\linewidth}{!}{
		\begin{tikzpicture}
		\pgfmathsetmacro{\fk}{2}
		\pgfmathsetmacro{\tk}{3*\fk}
		\pgfmathsetmacro{\nk}{1/(1/\fk-1/\tk)}
		\pgfmathsetmacro{\a}{2*\fk}
		\pgfmathsetmacro{\h}{0.5}
		\pgfmathsetmacro{\htk}{\tk/\nk*\h}
		\pgfmathsetmacro{\ha}{1.5*\h}
		\pgfmathsetmacro{\nl}{\nk-\h/(\h-\ha)*(\nk-\a)}
		\pgfmathsetmacro{\fn}{\nk-\h/(\h-\ha)*(\nk-\nl)}
		\coordinate (TK) at (\tk, -\htk);
		\coordinate (NK) at (-\nk, \h);
		\coordinate (A) at (-\a, \ha);
		\draw (-1.1*\fn, 0) -- (1.1*\tk, 0);
		\draw[<->, ultra thick] (0,3.5*\h) -- (0,-3.5*\h);
		\draw[->, ultra thick] (\tk, 0) -- (TK);
		\draw[fill] (-\fk, 0) circle (0.02*\fk) node[below] {$F_k$};
		\draw[fill] (\fk, 0) circle (0.02*\fk) node[below] {$F_k$};
		\draw[->, ultra thick] (-\a, 0) -- (A);
		\end{tikzpicture}}
\end{resizebox}

Nõguslääts tekitab esemest näiva kujutise ning kumerlääts tekitab näivast kujutisest tõelise kujutise. Kuna kumerläätse asukoht on teada, saab tagurpidi lähenedes rekonstrueerida näilise kujutise asukoha.


[User]
Consider a typical small LC circuit, with L = 10−7 H and C = 10−11 F.

     (a) If the circuit is at zero temperature, estimate the voltage uncertainty across the capacitor.

     (b) Do the same if the circuit is at room temperature.

    Today, some of the leading quantum computing hardware is based on such circuits. Because thermal
    fluctuations would ruin the desired quantum mechanical effects, the circuit must be cooled so that
    kB T ≪ ℏω, which corresponds to a small fraction of a degree. When you see pictures of quantum
    computers, most of what you’re looking at is the fridge!

Solution:
(a) In this problem, it’s not clear how to use Heisenberg’s uncertainty principle, because
         we don’t know what the “position” and “momentum” for an LC circuit are. (It turns out the
         position variable is the flux through the inductor and the momentum variable is the charge
         on the capacitor, but this is a bit hard to guess.) But we do know that an LC circuit behaves
         just like a harmonic oscillator, so when it is quantized, it has energy ℏω/2 in the ground state.
         This must manifest in terms of voltage and current uncertainties,

                                             ℏω ∼ C(∆V )2 + L(∆I)2

          where we drop all constants for brevity. We know from an example in X1 that for the harmonic
          oscillator, these two contributions are equal. Therefore, by order of magnitude we have
                                                   r
                                                     ℏω            1
                                            ∆V ∼         , ω=√        .
                                                      C            LC
          If you didn’t manage to do the steps above, this final result can also be found by pure
          dimensional analysis. Plugging in the numbers gives ∆V ∼ 10−7 V, which is surprisingly high!

     (b) At zero temperature, the energy in the circuit is of order ℏω, while at room temperature, it’s
         kB T by the equipartition theorem. Plugging in the numbers, we have
                                                 kB T
                                                      = 4 × 104 .
                                                  ℏω

         Since energy is proportional to V 2 , the voltage uncertainty is 200 times higher,

                                                ∆V ∼ 2 × 10−5 V.

         This is the typical noise you expect to see on a good oscilloscope. In electrical engineering,
         this kind of thermal noise is called “Johnson noise”.


[User]
Repeat problem 7 using the H field.
      Remark: Demagnetizing Fields

      At a given point in an isolated magnet, µ0 H is the part of B due to the rest of the magnet.
      However, the above example shows that H always points against the direction of M, so a
      permanent magnet is always trying to demagnetize itself! Similarly, electrically polarized
      materials carry an internal “depolarization” field. This is why, in the absence of external
      fields, the vast majority of materials have zero polarization and magnetization.

      Before the advent of very effective “hard” ferromagnets, like neodynium magnets, magnets
      had to be shaped to avoid this effect, e.g. by making them into long bars or horseshoes.
      Even so, the demagnetization effect would make the field produced by the magnet a little
      less than you would expect. For an average-shaped bar magnet made in the 1950s, the
      magnetic poles are effectively not at the ends, but rather 10% to 20% closer together.

      This bit of historical trivia is irrelevant today, but it has stuck around in the Indian physics
      curriculum. Every Indian introductory physics textbook demands its students memorize
      the ratio of the “magnetic length” and “geometric length” of a bar magnet, as if it were a
      fundamental constant of nature rather than an obsolete rule of thumb. But different ones
      don’t even agree on what the ratio is, with HC Verma giving 84%, various JEE prep sources
      stating 4/5, 5/6, or 7/8, and none whatsoever explaining where the number comes from.
      Such “magic formulas” are depressingly common in Indian books and should just be ignored.

      Idea 8: Magnetic Energy, Force, and Torque

      The appropriate magnetic energy density depends on the material.

        • If the magnetization is permanent, as in a hard ferromagnet, we should use the same
          potential energy introduced in E4. Specifically, the potential energy density is −M · B.

        • For all other materials, where the magnetization is induced by the presence of other fields,

           the total energy density is B 2 /2µ.

       If you use one of these formulas where the other applies, you’ll typically be off by a factor of 2.

       Regardless of how the magnetization arises, the resulting force and torque in an external field
       B are given by the formulas introduced in E4. Specifically, the torque density is M × B, and
       the force density is ∇(M · B), where the ∇ only acts on B. Alternatively, if you’re using the
       magnetic pole trick, a magnetic charge qm feels a force qm B.

Solution:
In the gap, the magnetization vanishes and we have B = µ0 H, so we can use the idea of
magnetic charge to compute the field. The magnetic charge is the same as that of a parallel plate
capacitor, with circular plates of radius r and separation 2a, carrying uniform magnetic charge
density ±M . Using Coulomb’s law, we can integrate the vertical component of the field from each
plate. If s is the distance of a point on the plate from the axis of symmetry, then
              Z r                                                r
                           µ0 M      a                     a                           a
      B=2         (2πs ds)                   = µ0 M − √               = µ0 M 1 − √             .
                0           4π (s2 + a2 )3/2             s2 + a2 0                  r 2 + a2
In the limit a ≪ r, this recovers the answer to problem 7.

   Remark: Demagnetizing Fields

   At a given point in an isolated magnet, µ0 H is the part of B due to the rest of the magnet.
   However, the above example shows that H always points against the direction of M, so a
   permanent magnet is always trying to demagnetize itself! Similarly, electrically polarized
   materials carry an internal “depolarization” field. This is why, in the absence of external
   fields, the vast majority of materials have zero polarization and magnetization.

   Before the advent of very effective “hard” ferromagnets, like neodynium magnets, magnets
   had to be shaped to avoid this effect, e.g. by making them into long bars or horseshoes.
   Even so, the demagnetization effect would make the field produced by the magnet a little
   less than you would expect. For an average-shaped bar magnet made in the 1950s, the
   magnetic poles are effectively not at the ends, but rather 10% to 20% closer together.

   This bit of historical trivia is irrelevant today, but it has stuck around in the Indian physics
   curriculum. Every Indian introductory physics textbook demands its students memorize
   the ratio of the “magnetic length” and “geometric length” of a bar magnet, as if it were a
   fundamental constant of nature rather than an obsolete rule of thumb. But different ones
   don’t even agree on what the ratio is, with HC Verma giving 84%, various JEE prep sources
   stating 4/5, 5/6, or 7/8, and none whatsoever explaining where the number comes from.
   Such “magic formulas” are depressingly common in Indian books and should just be ignored.

   Idea 8: Magnetic Energy, Force, and Torque

   The appropriate magnetic energy density depends on the material.

     • If the magnetization is permanent, as in a hard ferromagnet, we should use the same
       potential energy introduced in E4. Specifically, the potential energy density is −M · B.

     • For all other materials, where the magnetization is induced by the presence of other fields,
       the total energy density is B 2 /2µ.

   If you use one of these formulas where the other applies, you’ll typically be off by a factor of 2.

   Regardless of how the magnetization arises, the resulting force and torque in an external field
   B are given by the formulas introduced in E4. Specifically, the torque density is M × B, and
   the force density is ∇(M · B), where the ∇ only acts on B. Alternatively, if you’re using the
   magnetic pole trick, a magnetic charge qm feels a force qm B.


[User]
\begin{wrapfigure}[8]{r}{40mm}
	\vspace{-10pt}
	\includegraphics[width=40mm]{2011-lahg-10-kodar.pdf}
\end{wrapfigure}
Horisontaalsel pinnal veerevast radiaalsete kodaratega rattast tehakse pilt.
Fotokaamera säriaeg on mõõduka pikkusega: paigalseisvad objektid on pildil teravad,
liikuvad esemed aga hägused. Muuhulgas on ratta kodarad valdavalt hägused,
kuid osade kodarate teatud punktid on ometigi teravad. Võib eeldada, et kogu pilt on
salvestatud samaaegselt.
\\
\osa Kopeerige juuresolev skeem lahenduslehele ning näidake konstruktsiooni teel,
milline kodara $OP$ punkt (või punktid) kujutub fotol teravalt; põhjendage vastust.\\
\osa Konstrueerige kõver, millel asuvad ülejäänud kodarate teravalt kujutuvad punktid.

Solution:
Kodara antud punkt näib kujutisel terav, kui selle kiirusvektor on suunatud pikki kodarat, st antud punktis kodar ei liigu enese ristsihis.

\begin{wrapfigure}{r}{0.3\textwidth}
	\vspace*{-15pt}
	\includegraphics[width=0.3\textwidth]{2011-lahg-10-kodar_a}
	\vspace*{-25pt}
\end{wrapfigure}

Olukorda võib selgitada juuresoleva joonise abil.
Olgu $R$ ratta raadius ja olgu selle keskpunt $O$. Kui kodara pöördenurk on $\theta$ ning see nurk muutub pildistamise jooksul nurga $\D\theta$ võrra,
siis $O$ on läbinud teatud vahemaa ($R \, \D\theta$), aga kodar on samuti pöördunud sama nurga ($\D \theta$) võrra.
Jooniselt on näha, et uuel ja vanal kodara asendil on üks ühine punkt, olgu see punkt $Q$. Nii pildistamise alg- kui ka lõpphetkel asus selles punktis kodar, mistõttu kujutisel jääb see punkt selgelt näha (erinevalt teistest punktidest, kus kodar viibis vaid lühiajaliselt).

Kasutades eeltoodud joonist (kus tähistasime $OQ = r$) võime avaldada lõigu $MN$
pikkuse kahel viisil: $R \,\D \theta \cos \theta = r \,\D \theta$, kus
paremal pool kasutasime väikese nurga lähendust $\sin \D\theta\approx \D\theta$. Seega $R \cos \theta = r$, mis tähendab, et
(a) punkt $Q$ on leitav kodara lõikepunktina ratta ja maa kontaktpunktist $S$ kodarale tõmmatud ristsirgega (vt järgnev joonis);
(b) vaadeldes seda võrdust kui raadiuse $r$ sõltuvust polaarnurgast $\theta$ näeme, et ülejäänud kodarate teravalt kujutuvad punktid asuvad ringjoonel,
mille diameetriks on ratta raadius $OS$.

\vspace{0.5\baselineskip}

\textit{Alternatiivne lahendus}

Pildistamise hetkel pöörleb kogu ratas ümber hetkelise pöörlemistelje, mis läbib
ratta ja maa puutepunkti $S$ (vt. joonist). Sellel hetkel liigub iga ratta osake
mööda ringjoone kaart, mille keskpuntiks on $S$. Kui ühe sellisel moel liikuva
punkti kiirus on mööda kodarat ($OP$), siis see punkt kujutub fotol selgena.
Seega me otsime selliseid punkte $Q$, mille juures $\angle OQS$ on täisnurk.
Piirdenurga omaduse põhjal peab selline punkt $Q$ lebama ringjoonel, mille
diameetriks on $OS$ --- sõltumata kodara kaldenurgast $\theta$.
\begin{center}
	\includegraphics[width=0.65\textwidth]{2011-lahg-10-kodar_b}
\end{center}

Kodarad

\begin{wrapfigure}[8]{r}{40mm}
	\vspace{-10pt}
	\includegraphics[width=40mm]{2011-lahg-10-kodar.pdf}
\end{wrapfigure}
Horisontaalsel pinnal veerevast radiaalsete kodaratega rattast tehakse pilt.
Fotokaamera säriaeg on mõõduka pikkusega: paigalseisvad objektid on pildil teravad,
liikuvad esemed aga hägused. Muuhulgas on ratta kodarad valdavalt hägused,
kuid osade kodarate teatud punktid on ometigi teravad. Võib eeldada, et kogu pilt on
salvestatud samaaegselt.
\\
\osa Kopeerige juuresolev skeem lahenduslehele ning näidake konstruktsiooni teel,
milline kodara $OP$ punkt (või punktid) kujutub fotol teravalt; põhjendage vastust.\\
\osa Konstrueerige kõver, millel asuvad ülejäänud kodarate teravalt kujutuvad punktid.

Pildistamise hetkel pöörleb kogu ratas ümber hetkelise pöörlemistelje, mis läbib
ratta ja maa puutepunkti. See tähendab, et ratta iga osake liigub mööda ringjoone kaart, mille keskpunktiks on ratta ja maa puutepunkt.


[User] Õhukeste seintega jääst kera sees on õhk. Algselt on jääkera külmkapis temperatuuril $t_0 = \SI{-9}{\degreeCelsius}$ ning õhurõhk tema sees võrdub välisrõhuga $p_0 = \SI{105}{kPa}$. Kera tõstetakse külmikust välja tuppa, kus see hakkab soojenema. Kera sein on nii õhuke, et maksimaalne ülerõhk (st. sise- ja välisrõhkude vahe), mida ta purunemata talub, on $\Delta p = \num{0,2}p_0$. Mis juhtub enne: kas kera hakkab sulama või ta puruneb ülerõhu tõttu? Kuuli soojenemine lugeda nii aeglaseks, et igal ajahetkel võib lugeda õhu temperatuuri tema sees ning seinte sise- ja välispinna temperatuurid võrdseks.

Solution:
Rõhk kuuli sees kasvab seetõttu, et õhk kuulis soojeneb. Ülesande teksti põhjal võime eeldada, et õhu temperatuur kuuli sees on võrdne tema seinte temperatuuriga. Meie ülesandeks on kontrollida, kui palju on rõhk kasvanud selleks hetkeks, kui seinad hakkavad sulama, st on saavutanud temperatuuri $t_1 = \SI{0}{\degreeCelsius}$.

Eeldame, et kera soojuspaisumine on tühine. Siis on õhu ruumala keras konstantne. Isohoorilises protsessis kehtib seos
\[
\frac{p_1}{T_1} = \frac{p_0}{T_0}.
\]
Tähistame indeksiga \enquote{0} gaasi omadusi külmikus ja indeksiga \enquote{1} omadusi temperatuuril, mille juures seinad hakkavad sulama. Niisiis
\[
p_1 = p_0 \frac{T_1}{T_0}.
\]
Kasutades seda tulemust saame avaldada rõhu suhtelise muutuse
\[
\frac{\Delta p}{p_{0}}=\frac{p_{1}-p_{0}}{p_{0}}=\left(\frac{T_{1}}{T_{0}}-1\right).
\]
Leiame selle avaldise numbrilise väärtuse:
\[
\frac{\Delta p}{p_0} \approx \num{0,034} = \SI{3,4}{\%}.
\]
See on selgelt väiksem kui kuuli seinte purunemispiir, st kuul hakkab enne sulama (kuid puruneb ilmselt ülerõhu tõttu enne lõplikku ära sulamist).

\emph{Märkus:} alternatiivse ja võrdväärse lahendusena võib leida, millise õhutemperatuuri juures saavutaks suhteline ülerõhk väärtuse \SI{20}{\%} (selleks tuleb \SI{317}{K} ehk \SI{44}{\degreeCelsius}) ja võrrelda seda jää sulamistemperatuuriga.

Jääkuul

Õhukeste seintega jääst kera sees on õhk. Algselt on jääkera külmkapis temperatuuril $t_0 = \SI{-9}{\degreeCelsius}$ ning õhurõhk tema sees võrdub välisrõhuga $p_0 = \SI{105}{kPa}$. Kera tõstetakse külmikust välja tuppa, kus see hakkab soojenema. Kera sein on nii õhuke, et maksimaalne ülerõhk (st. sise- ja välisrõhkude vahe), mida ta purunemata talub, on $\Delta p = \num{0,2}p_0$. Mis juhtub enne: kas kera hakkab sulama või ta puruneb ülerõhu tõttu? Kuuli soojenemine lugeda nii aeglaseks, et igal ajahetkel võib lugeda õhu temperatuuri tema sees ning seinte sise- ja välispinna temperatuurid võrdseks.

Jääkera sees tekib ülerõhk, sest õhu soojenedes rõhk tõuseb. Seega on kõige kriitilisem moment vee sulamise temperatuuril.


[User]
Risti üle $l=\SI{100}{m}$ laiuse jõe kulgeb kumer autoteega sild. Silla keskel on autotee kaldapealsest tasemest $h=\SI{5}{m}$ võrra kõrgemal. Silla profiiliks on
ringjoone kaar. Auto massiga $m=\SI{1000}{kg}$ ületab silda muutumatu kiirusega $v=\SI{60}{km/h}$.
Kui suure jõuga rõhub auto silla keskkohta? Kui suure kiiruse juures hakkab kaduma
kontakt rataste ja tee vahel?

Solution:
Olgu
silla kõverusraadius $r$. Pythagorase teoreemist
\[
r^2=(l/2)^2+(r-h)^2\implies 0=l^2/4-2rh+h^2.
\]
Kuna $h\ll l$ ja seega $h\ll r$, siis $h^2$ võib ära jätta ja $r=l^2/8h=\SI{250}{m}$.
Auto raskusjõu $mg$ ja toereaktsiooni $N$ resultant annab kesktõmbekiirenduse $v^2/r$.
Seega $N=mg-mv^2/r\approx\SI{8700}{N}$.

Kontakt rataste ja maapinna vahel hakkab kaduma, kui $N=0$. Seega $v=\sqrt{gr}\approx
\SI{180}{km/h}$.

Sild

Risti üle $l=\SI{100}{m}$ laiuse jõe kulgeb kumer autoteega sild. Silla keskel on autotee kaldapealsest tasemest $h=\SI{5}{m}$ võrra kõrgemal. Silla profiiliks on
ringjoone kaar. Auto massiga $m=\SI{1000}{kg}$ ületab silda muutumatu kiirusega $v=\SI{60}{km/h}$.
Kui suure jõuga rõhub auto silla keskkohta? Kui suure kiiruse juures hakkab kaduma
kontakt rataste ja tee vahel?

Silla kõverusraadius on leitav Pythagorase teoreemist. Autole mõjub silla peal kaks jõudu: raskusjõud ja rõhumisjõud. Antud jõudude resultant annab kesktõmbekiirenduse.


[User] Kuubik massiga $m = \SI{10}{kg}$ ning küljepikkusega $a = \SI{0,1}{m}$ lebab laual. Laua ja kuubiku vaheline hõõrdetegur on $\mu = \num{0,5}$. Kas kuubikut on võimalik käega teisele küljele lükata, avaldades vaid jõudu kuni $F = \SI{40}{N}$? Eeldada, et hõõrdetegur käe ja kuubiku vahel on väga suur ehk käsi ei libise. Raskusjõu kiirendus on $g = \SI{9,8}{m/s^2}$.

Solution:
Ülesande lahendamine jaguneb kaheks osaks: (a) kas antud jõust piisab üle serva kantimiseks; (b) ega klots seejuures libisema ei hakka. Analüüsides oletame, et klots on juba kallutatud teatud nurga $\varphi$ ($0 \leq \varphi \leq \ang{45}$) võrra; seejuures selgub, et $\varphi = 0$ on kõige ohtlikum olukord. Alternatiiv oleks väita intuitiivselt, et ohtlikuim on olukord $\varphi = 0$ ning uurida ainult seda juhtumit;

\begin{center}
	\includegraphics[height=0.6\textheight]{2007-v2g-09-lah}
\end{center}

(a) Vaatleme jõumomentide tasakaalu toetava nurga suhtes. Kompenseerimist vajab raskusjõu moment $M_{1 \mathrm{max}} = F a \cos (\ang{45}) \cos (\varphi + \ang{45})$, mille maksimaalväärtus on
\[
M_{1 \max }=\frac{m g a}{2}, \quad M_{1 \max }=\frac{10 \cdot \num{9,8} \cdot 0,1}{2}=\SI{4,9}{N.m}.
\]
Rakendatav jõud annab seda suurema momendi, mida suurem on õlg; õla maksimaalne pikkus ei sõltu nurgast $\varphi$ ning on alati $l = a \sqrt 2$. See väärtus saavutatakse siis, kui jõud on rakendatud maha toetuva serva suhtes vastasserva külge ning on risti ruudu diagonaaliga. Seega on antud jõu abil alati võimalik tekitada raskusjõudu kompenseeriv moment väärtusega kuni
\[
M_{2}=F l=F a \sqrt{2} = \SI{5.6}{N.m}.
\]
Näeme, et $M_{1 \mathrm{max}} < M_2$, st antud jõud on piisav kuubi keeramiseks.

(b) Vaatleme jõudude tasakaalu raskusjõu $m\vec g$ ja rakendatud jõu $\vec F$ pikenduste lõikepunkti $O$ suhtes, vt joonist. Aeglasel pööramisel on jõud tasakaalus, st hõõrdejõu ja toereaktsiooni resultantjõud $\vec f$ peab minema samuti läbi selle punkti. Et hõõrdetegur $\mu = \num{0,5}$, siis nurk toetuspinna normaali (st vertikaalsihi) ja jõu $\vec f$ vahel ei tohi olla suurem, kui $\arctan \mu$, st jõud $\vec f$ ei tohi olla vähem püstine, kui sirges. Nii see ka tõepoolest on, sest punkt $O$ jääb alati piirkonda $x \leq 0$ ja $y > 0$.

\vspace{0.5\baselineskip}

\emph{Alternatiivne lahendus osa (b) jaoks}

Meil on vaja tõestada, et aeglasel pööramisel kehtib kogu aeg võrratus
\[
\left|F_{x}\right|=F \cos \left(\ang{45}-\varphi\right) \leq N \mu,
\]
kus $N$ on laua toereaktsioon. Paneme tähele, et vertikaalsest tasakaalutingimusest
\[
N=m g-\left|F_{y}\right|=m g-F \sin \left(\ang{45}-\varphi\right).
\]
Me kasutame osast (a) teada olevat asjaolu, et kui hõõrdumist ei oleks, siis tasakaalu tagava jõu jaoks kehtib võrratus $F < F\idx{max}$, seda asjaolu kasutame alljärgnevalt võrratuste ümber kirjutamisel.

Meile piisaks, kui suudaksime tõestada, et
\begin{equation}\label{2007-v2g-08:eq1}
\mu [mg - F\idx{max} \sin (\ang{45} - \varphi )] \geq F\idx{max} \cos (\ang{45} - \varphi ),
\end{equation}
sest sellisel juhul
\[
\begin{aligned}
N \mu &=\mu\left[m g-F \sin \left(45^{\circ}-\varphi\right)\right] \geq \mu\left[m g-F_{\max } \sin \left(\ang{45}-\varphi\right)\right] \geq \\ & \geq F_{\max } \cos \left(\ang{45}-\varphi\right) \geq F \cos \left(\ang{45}-\varphi\right)=\left|F_{x}\right|.
\end{aligned}
\]
Tõepoolest, $N\mu \geq |F_x|$. Võrratuse (\ref{2007-v2g-08:eq1}) tõestamiseks kirjutame selle ümber ekvivalentsel kujul
\[
1 \geq \frac{F_{\max }}{\mu m g}\left[\mu \sin \left(\ang{45}-\varphi\right)+\cos \left(\ang{45}-\varphi\right)\right],
\]
mis tõepoolest kehtib, sest
\[
\begin{aligned}
& \frac{F_{\max }}{\mu m g}\left[\mu \sin \left(\ang{45}-\varphi\right)+\cos \left(45^{\circ}-\varphi\right)\right]=\\=& \frac{F_{\max } \sqrt{\mu^{2}+1}}{\mu m g} \sin \left(45^{\circ}-\varphi+\arcsin \left[\left(\mu^{2}+1\right)^{-1}\right]\right) \leq \\ \leq & \frac{F_{\max } \sqrt{\mu^{2}+1}}{\mu m g}=\frac{\SI{40}{N} \cdot \sqrt{5 / 4}}{\SI{49}{N}} \approx \num{0,91}<1.
\end{aligned}
\]

Kuubik

Kuubik massiga $m = \SI{10}{kg}$ ning küljepikkusega $a = \SI{0,1}{m}$ lebab laual. Laua ja kuubiku vaheline hõõrdetegur on $\mu = \num{0,5}$. Kas kuubikut on võimalik käega teisele küljele lükata, avaldades vaid jõudu kuni $F = \SI{40}{N}$? Eeldada, et hõõrdetegur käe ja kuubiku vahel on väga suur ehk käsi ei libise. Raskusjõu kiirendus on $g = \SI{9,8}{m/s^2}$.

Ülesande lahendamiseks tuleb küsida: (a) kas antud jõust piisab üle serva kantimiseks; (b) ega klots seejuures libisema ei hakka? Mõlema küsimuse analüüsimiseks on mugav rakendada jõumomentide tasakaalu. Lisaks on kasulik teada, et hõõrdejõu ja toereaktsiooni resultantjõu maksimaalne nurk vertikaali suhtes on $\arctan \mu$.


[User]
7. As shown, a magnet of mass m = 19.00g slides on a rough
   non-magnetic metallic inclined plane which makes an angle θ
   with the horizontal. One can change the angle of inclination
   of the plane. Due to relative motion between the magnet
   and the plane, eddy currents are generated inside the metal
   which retard motion of the magnet.
   Assume that the magnitude of the resistive force on the magnet is bv where b is a positive
   constant and v is the instantaneous speed of the magnet. Let µ be the coefficient of kinetic
   friction between the magnet and the plane.
   (a) If the magnet starts moving from rest at time t = 0, obtain the expression of its terminal   [2]
       velocity VT . Also obtain the displacement S(t) along the inclined plane as a function of

INPhO 2019 Page 8        Questions & Summary Answers                Last four digits of Roll No.:
       time. Take S(0) = 0.

        VT =

        S(t) =

   (b) For a fixed θ a student records S for all values of t as shown in the table below. Draw      [4]
       a suitable graph and obtain the terminal velocity VT of the magnet from this graph. For
       this and the next part, three graph papers are provided with this booklet. No extra graph
       papers will be provided.
                                             t (s) S (m)
                                             0.016 0.001
                                             0.049 0.003
                                             0.070 0.006
                                             0.090 0.010
                                             0.120 0.017
                                             0.174 0.029
                                             0.230 0.046

                                  C1                9
                                             0.270 0.058
                                             0.320 0.074
                                             0.370 0.091

        VT =

       Graph is plotted on page no. : ____

               HB
    (c) The above process is repeated for various values of θ. The obtained values of terminal
        velocities for the different θ are given below.
                                              θ (degree) VT (m/s)
                                                  19
                                                  24
                                                  28
                                                           0.15
                                                           0.23
                                                           0.29
                                                                                [7]

                                                  35       0.40
                                                  40       0.46
                                                  45       0.53
                                                  48       0.58
                                                  52       0.62
        Plot a suitable graph and obtain the values of b and µ from this graph.

        b=                                         µ=

       Graph is plotted on page no. : ____
                              Detailed answers can be found on page numbers:

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

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Solution:
7. As shown, a magnet of mass m = 19.00g slides on a rough
   non-magnetic metallic inclined plane which makes an angle θ
   with the horizontal. One can change the angle of inclination
   of the plane. Due to relative motion between the magnet
   and the plane, eddy currents are generated inside the metal
   which retard motion of the magnet.
   Assume that the magnitude of the resistive force on the magnet is bv where b is a positive
   constant and v is the instantaneous speed of the magnet. Let µ be the coefficient of kinetic
   friction between the magnet and the plane.
   (a) If the magnet starts moving from rest at time t = 0, obtain the expression of its terminal       [2]
       velocity VT . Also obtain the displacement S(t) along the inclined plane as a function of
       time. Take S(0) = 0.

        VT =

        S(t) =

INPhO 2019 Page 17             Questions & Answers

         Solution:
                                          mg(sin θ − µ cos θ)
                                     VT =                                                (7.1)
                                                   b
                                                  m

                                                           −tb/m
                                   S(t) = VT t −     1−e                                 (7.2)
                                                  b

   (b) For a fixed θ a student records S for all values of t as shown in the table below. Draw     [4]
       a suitable graph and obtain the terminal velocity VT of the magnet from this graph. For
       this and the next part, three graph papers are provided with this booklet. No extra graph
       papers will be provided.
                                             t (s) S (m)
                                             0.016 0.001
                                             0.049 0.003
                                             0.070 0.006
                                             0.090 0.010
                                             0.120 0.017
                                             0.174 0.029
                                             0.230 0.046

        VT =                             C1
                                             0.270 0.058
                                             0.320 0.074
                                             0.370 0.091
                                                           9
       Graph is plotted on page no. : ____

                     HB
         Solution: VT = slope of the graph = 0.315 m/s
         Accepted range 0.301 m/s ≤ VT ≤ 0.329 m/s.

    (c) The above process is repeated for various values of θ. The obtained values of terminal
        velocities for the different θ are given below.
                                              θ (degree) VT (m/s)
                                                                                [7]

                                                  19       0.15
                                                  24       0.23
                                                  28       0.29
                                                  35       0.40
                                                  40       0.46
                                                  45       0.53
                                                  48       0.58
                                                  52       0.62
        Plot a suitable graph and obtain the values of b and µ from this graph.

        b=                                         µ=

       Graph is plotted on page no. : ____

INPhO 2019 Page 18           Questions & Answers                 Last four digits of Roll No.:

        Solution:
                                      VT     mg         µmg
                                           =    tan θ −                                (7.3)
                                     cos θ    b          b
        Graph is plotted for VT / cos θ vs tan θ.
                                          tan θ VT / cos θ (m/s)
                                          0.34        0.16
                                          0.45        0.25
                                          0.53        0.33
                                          0.70        0.49
                                          0.84        0.60
                                          1.00        0.75
                                          1.11        0.87
                                          1.28        1.01
                                           mg
        Slope of the graph = 0.91 m/s =        ⇒ b = 0.21 N·s/m
                                            b
                           µmg
        Intercept = 0.16 =        ⇒ µ = 0.18
                             b
        Accepted range of slope = 0.86 - 0.96 m/s
        Accepted range of intercept = 0.12 - 0.19 m/s
        Respective ranges of b and µ are
        0.19 ≤ b ≤ 0.22 m/s and 0.14 ≤ µ ≤ 0.20

                                C1               9
        Determination of b and µ from a graph of VT vs θ using an appropriate method also
        gains credit.

                            Detailed answers can be found on page numbers:

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INPhO 2019 Problem 7


[User]
8. Loop of wire (7 points)

Consider a rectangular loop of wire with dimen-
sions a = 0.03m and b = 1.0m, one side
of which is parallel to another long straight wi-
re carrying current I0 = 1000A, at distance
l = 0,01m,seefigure. Themagneticinductance
ofsuchcurrentisplottedasafunctionofthedis-
tance from the wire in attached graph. The Oh-
mic resistance of the loop is R = 1,0Ω, the in-
ductance is negligible.
b
b
I0
l a
b
1) Calculate the magnetic flux Φ through the
loop (2 pts).
2) At a certain moment of time, the current in
the long wire is switched off. What is the net
charge Q flowing through a fixed cross-section
of the wire of the loop (3 pts)?
3) What is the net momentum p given to the
loopduringtheswitch-offofthecurrent(express
it in terms of Q and the given quantities, if you
were unable to calculate Q) (2 pts)?

Solution:
8. Loop of wire

1) At the distance r from the current I0, the
magnetic induction B = μ0I0
2πr . Then, the
flux through the contour Φ =
Rl+a
l Bbdx =
Rl+a
l
bμ0I0
2πx dx = bμ0I0
2π ln a+l
l .
Alternatively, we can find it using the graph
by determining the area S under the curve, from
r = r1 = 0,01m to r = r2 = 0,04m:
S ≈ 0,28mT·m, further,Φ = Sb = 280μWb.
2) After switching off the current, the flux th-
rough the tends to zero. From the Ohm’s law
Rdq
dt = dΦ
dt , hence Rdq = dΦ, i.e. RQ = ∆Φ =
Φ. Finally, Q = Φ/R = 280μC.
3) We calculate the force as difference between
the forces at the two loop segments parallel
to the straight line: F1 = biB1 and F2 =
biB2, where i = R−1 dΦ
dt . So, dp = (F1 −
F2)dt = bR−1
(B1 − B2)dΦ. Using B1 =
μ0I/2πl and B1 = μ0I/2π(l + a), we end
up with dp = b
R
μ0I
2π
a
l(l+a)dΦ. Using the re-
sult of first question, dΦ = bμ0
2π ln a+l
l dI, i.e.
dp = (bμ0
2π )2 a
Rl(l+a) ln a+l
l IdI. Finally, p =
a(bμ0I0)2
8π2Rl(l+a) ln a+l
l ≈ 2.08 · 10−6
kg·m/s2
.
The same result could have been obtained
using the graph and approach used in the alter-
native solution of the question 1.

Loop of wire


[User]
Toomas mängib läbipaistvast aiavoolikust tehtud U-toruga. Et seekordne U-toru polegi klaasist, painutab ta üht poolt nurga $\alpha$ ning teist $\beta$ võrra (vt joonist). Kas vedelikutaseme võnkesagedus on nüüd suurem või väiksem, mitu korda?

\emph{Märkus:} vertikaalses U-torus on vedelikutaseme võnkumise sagedus $f = \frac{1}{2\pi} \sqrt{\frac{2S\rho g}{m}}$, kus $S$ on toru ristlõikepindala, $\rho$ vedeliku tihedus ning $m$ torus oleva vedeliku mass.

\begin{center}
	\includegraphics[width=0.5\linewidth]{2008-v3g-08-yl}
\end{center}

Solution:
Oletame, et vesi on tasakaalust hälbinud nii, et veekiht, mille kõrguse projektsioon toru sihis on $\delta x$, on kandunud nurga $\alpha$ all olevast toru poolest teise. Veetaseme kõrguste erinevus toru kahe poole vahel on $\delta h = \delta x (\cos (\alpha ) + \cos (\beta ))$, mis tekitab lisarõhu $\delta P = \rho \delta hg$. Lisarõhk mõjub vedelikule jõuga, mille toru sihiline komponent on
\[
F = \delta P S = \rho \delta x (\cos (\alpha ) + \cos (\beta )) gS.
\]
See valem on sarnane vedrupendli valemiga $F = k\delta x$, kus
\[
k = \rho (\cos (\alpha ) + \cos (\beta )) gS.
\]
Sellise pendli omavõnkesagedus avaldub kui
\[
f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} = \frac{1}{2\pi} \sqrt{\frac{\rho(\cos(\alpha)+\cos(\beta))gS}{m}}.
\]
Seega võime öelda, et toru pooli painutades väheneb vedeliku võnkumise sagedus
\[
\sqrt{\frac{1}{2}(\cos \alpha + \cos \beta)}
\]
korda.

V-toru

Toomas mängib läbipaistvast aiavoolikust tehtud U-toruga. Et seekordne U-toru polegi klaasist, painutab ta üht poolt nurga $\alpha$ ning teist $\beta$ võrra (vt joonist). Kas vedelikutaseme võnkesagedus on nüüd suurem või väiksem, mitu korda?

\emph{Märkus:} vertikaalses U-torus on vedelikutaseme võnkumise sagedus $f = \frac{1}{2\pi} \sqrt{\frac{2S\rho g}{m}}$, kus $S$ on toru ristlõikepindala, $\rho$ vedeliku tihedus ning $m$ torus oleva vedeliku mass.

\begin{center}
	\includegraphics[width=0.5\linewidth]{2008-v3g-08-yl}
\end{center}

Omavõnkesageduse leidmiseks on harilikult kõige mugavam vaadelda väikest hälvet tasakaaluasendist ning uurida, kuidas süsteem edasi käitub. Ülesande kontekstis võib oletada, et õhuke veekiht ühes toru harus kandub teisele toru poolele. See põhjustab lisarõhu toru teises pooles, mis üritab süsteemi tasakaaluasendisse tagasi viia.


[User]
3. Tractor (6 points)

Provided sketches (a) and (b) are made on the basis of satel-
liteimages,preservingproportions. Theyrepresenttractors,to-
gether their smoke trails. The tractors were moving along the
roadsinthedirectionindicatedbythearrows. Thevelocityof
allthetractorswasv0 = 30km/h. Forsketch(a),thedirection
ofwindisindicatedbyanotherarrow.
1) Usingtheprovidedsketch,findthewindspeedforcase(a).
2) Usingtheprovidedsketch, findthewindspeedforcase(b).
(a)
(b)

Solution:
3. Tractor

1) LetusdrawfromanarbitrarypointB ontheroadalinepar-
alleltothedirectionofthewind,andletitintersectthesmoke
trailatpointC. Then, thesmokeemittedbythetractoratB
hastravelledthedistance|BC| = ut,whereu isthewindspeed.
Tractoritselfhastravelledthedistance|AB| = v0t. So,wecan
measurethedistances|AB| and|BC fromthefigureandcalcu-
late
u = v0
|BC|
|AC|
=
18mm
42mm
30km/h ≈ 13km/h.
2) Ifthesecondtractor(attheright-hand-side)hadstartedsome-
whatearlier, thetwotractorshadbeenatthecrossroadsimul-
taneously. Now,thetractorswouldbeatthesamedistancefrom
thecrossroad,i.e.forthecurrentpositionofthesecondtractor
C, |OC| = |AO| = v0t (this is how we find the point C). Its
smoketrailcanbefoundasaline,paralleltoitssmoketrailat
itsactualpositionB. Suchameetingofthetractorswouldhave
beenresultedinthecrossingofthesmoketrails. whichwouldbe
nowinpositionD,withOD = ut. So,wefind
u = v0
|OD|
|AO|
=
27mm
39mm
30km/h ≈ 21km/h.

Tractor


[User]
\begin{wrapfigure}[8]{r}{0.3\textwidth}
	\begin{center}
		\vspace{-25pt}
		\includegraphics[width=\linewidth]{2006-v3g-10-yl}
	\end{center}
\end{wrapfigure}
Joonisel kujutatud süsteem koosneb kolmest võrdkülgse kolmnurga tippudes paiknevast kuulist massiga $m$ ja kolmest kergest vardast pikkusega $l$, mis on omavahel šarniirselt ühendatud (liigendiga). Süsteem lebab hõõrdevabalt siledal horisontaalpinnal. Ühte kuuli lükatakse teatud lühiajalise jõuga nii, et see omandab kiiruse $v_0$, mis on suunatud naaberkuuli poole. Leidke teiste kuulide kiiruste suunad ja moodulid ning kõigi kuulide kiirendused vahetult peale esimese kuuli lükkamist.

Solution:
Teisest ja kolmandast kuulist koosnevale süsteemile mõjus sel ajal, kui esimest kuuli lükati, esimese varda sihiline jõud, sest teatavasti mõjuvad kergetele varrastele vaid varda sihilised pinged. Seega nihkus šarniirne ühenduspunkt esimese varda sihiliselt ning teine ja kolmas kuul omandasid sümmeetria tõttu ühesugused kiirused. Šarniirse ühenduspunkti kiirus on $v_s = v_0 \cos \ang{30}$, sest esimese varda pikkus ei muutu. Et nii teisele kui kolmandale kuulile mõjub ainult varda sihiline jõud, siis nende kiirus on ka varda sihiline; varda venimatusest juhtivalt $v_2 = v_3 = v_s \cos \ang{60} = v_0 \sqrt 3/4$.

Pinged varrastes on võrdsed, sest šarniirse ühenduspunkti massi võime lugeda nulliks ning talle mõjuv resultantjõud peab olema 0 ja jõudude tasakaalust tuginevalt peab varraste pingetest moodustuma võrdkülgne kolmnurk. Olgu varraste pinge $T$. Seega on kõigi kuulikeste kiirendused võrdsed, $a_k = T /m$. Läheme šarniirse ühenduspunktiga seotud kulgevalt liikuvasse taustsüsteemi, mis liigub kiirendusega $\vec a$. Selles süsteemis on esimese kuuli kiirus $u_1 = v_0/2$ ning teiste kiirus $u_2 = v_s \sin \ang{60} = \frac{3}{4} v_0$. Selles süsteemis liiguvad kuulid ringjoont mööda. Iga kuuli jaoks saame välja kirjutada jõudude tasakaalu tingimuse projekteerituna vastava varda sihile (siis kaob vajadus teada kuulikese ringliikumise tangetsiaalkiirendust, sest see on teljega risti). Teise ja kolmanda kuuli tasakaalutingimusi võrreldes selgub, et inetrsijõu $-m\vec a$ projektsioon kummalegi teljele peab olema üks ja sama, st $\vec a$ peab olema esimese varda sihiline. Seega saame kaks võrrandit:
\[
\begin{aligned}
mu_1^2/l + a &= T,\\
mu_2^2/l - a/2 &= T.
\end{aligned}
\]
Elimineerides neist võrrandeist $a$, saame
\[
T=\frac{1}{3} \frac{m}{l}\left(u_{1}^{2}+2 u_2^{2}\right)
\]
ning otsitava kiirenduse
\[
a_{k}=T / m=\left(u_{1}^{2}+2 u_2^{2}\right) 3 l=\frac{11}{24} v_{0}^{2} / l.
\]

Kuulid

\begin{wrapfigure}[8]{r}{0.3\textwidth}
	\begin{center}
		\vspace{-25pt}
		\includegraphics[width=\linewidth]{2006-v3g-10-yl}
	\end{center}
\end{wrapfigure}
Joonisel kujutatud süsteem koosneb kolmest võrdkülgse kolmnurga tippudes paiknevast kuulist massiga $m$ ja kolmest kergest vardast pikkusega $l$, mis on omavahel šarniirselt ühendatud (liigendiga). Süsteem lebab hõõrdevabalt siledal horisontaalpinnal. Ühte kuuli lükatakse teatud lühiajalise jõuga nii, et see omandab kiiruse $v_0$, mis on suunatud naaberkuuli poole. Leidke teiste kuulide kiiruste suunad ja moodulid ning kõigi kuulide kiirendused vahetult peale esimese kuuli lükkamist.

Teisest ja kolmandast kuulist koosnevale süsteemile mõjus esimese kuuli lükkamise ajal esimese varda sihiline jõud, sest teatavasti mõjuvad kergetele varrastele vaid varda sihilised pinged. Lisaks peab esimese kuuli lükkamise ajal kehtima varraste venimatuse tingimus.

Kiirenduse leidmiseks on süsteemi mugav vaadelda šarniirse ühenduspunktiga kaasa liikuvas ja kiirenevas taustsüsteemis ning seejärel rakendada Newtoni II seadust.


[User]
IPhO2002

                    III. A Heavy Vehicle Moving on An Inclined Road

                                                            L
                   m2                    m1
                rear cylinder                       2l

                                          m1

                                                                        t
                                                                    h
                                               m3
                                           front cylinder

                                                                è

        Figure III-1: A simplified model of a heavy vehicle moving on an inclined
        road.

        The above figure is a simplified model of a heavy vehicle (road roller) with
one rear and one front cylinder as its wheels on an inclined road with inclination angle
of è as shown in Figure III-1. Each of the two cylinders has a total mass
M(m2 =m3=M) and consists of a cylindrical shell of outer radius Ro , inner radius Ri =
0.8 Ro and eight number of spokes with total mass 0.2 M. The mass of the
undercarriage supporting the vehicle’s body is negligible. The cylinder can be
modeled as shown in Figure III-2. The vehicle is moving down the road under the
influence of gravitational and frictional forces. The front and rear cylinder are
positioned symmetrically with respect to the vehicle.

                                               6

IPhO2002

                                                                  Ri

                                                                              Ro

                                   Figure III-2: A simplified model of the cylinders.

The static and kinetic friction coefficients between the cylinder and the road are μs
and μk respectively. The body of the vehicle has a mass of 5M , length of L and
thickness of t . The distance between the front and the rear cylinder is 2l while the
distance from the center of cylinder to the base of the vehicle’s body is h. Assume that
the rolling friction between the cylinder and its axis is negligible.

Questions:
    1. Calculate the moment of inertia of either cylinder [1.5 pts].

    2. Draw all forces that act on the body, the front cylinder, and the rear one. Write
        down equations of motion for each part of them [2.5 pts].

    3. The vehicle is assummed to move from rest, then freely move under
        gravitational influence. State all the possible types of motion of the system and
        derive their accelerations in terms of the given physical quantities [4.0 pts].

    4. Assume that after the vehicle travels a distance d by pure rolling from rest the
        vehicle enters a section of the road with            all the friction coefficients drop to
        smaller constant values μs’ and μk ’ such that the two cylinders start to slide.
        Calculate the linear and angular velocities of each cylinder after the vehicle
        has traveled a total distance of s meters. Here we assume that d and s is much
        larger than the dimension of vehicle [2.0 pts]

                                                  6

IPhO2002

           6

Solution:
SOLUTION T3 : . A Heavy Vehicle Moving on An Inclined Road

                                       l

                                                       l

                                                                   h1

                                                       θ

 To simplify the model we use the above figure with h1 = h+0.5 t
 Ro = R

  1. Calculation of the moment inertia of the cylinder

      Ri=0.8 Ro
      Mass of cylinder part : mcylinder =0.8 M
      Mass of each rod       : mrod = 0.025 M

                                        III - 1

                2                   2              2                      2
I=            ∫ r dm = ∫ r dm + ∫ r dm + ... + ∫ r dm                                    0.4 pts
        wholepart         cyl . shell       rod1                  rod n
                          Ro
       2            3             4     4                   2     2
    ∫ r dm = 2πσ ∫ r dr = 0.5πσ( Ro − Ri ) = 0.5mcylinder (Ro + Ri )
cyl . shell               Ri

                    = 0.5(0.8 M ) R 2 (1 + 0.64) = 0.656 MR 2                            0.5 pts
               Rin
      2             2     1 3 1            2    1              2              2
∫rodr   dm = λ  ∫0 r dr = 3 λRin = 3 mrod Rin = 3 0.025M (0.64R ) = 0.00533MR            0.5 pts

The moment inertia of each wheel becomes
I = 0.656MR 2 + 8 x 0.00533MR 2 = 0.7 MR 2                                                0.1 pts

2. Force diagram and balance equations:

To simplify the analysis we devide the system into three parts: frame (part1) which
mainly can be treated as flat homogeneous plate, rear cylinders (two cylinders are treated
collectively as part 2 of the system), and front cylinders (two front cylinders are treated
collectively as part 3 of the system).

Part 1 : Frame

                N12
                                        l

      f12h                                                    l

                                                                  N13
                                  m1 g
                                                       f13h

                                                                              h1

                                                                                0.4 pts
The balance equation related to the forces work to this parts are:

                                                              III - 2

 Required conditions:
 Balance of force in the horizontal axis
 m1 g sin θ − f 12 h − f 13 h= m1 a                              (1) 0.2 pts
 Balance of force in the vertical axis
 m1 g cos θ = N 12 + N13                                        (2) 0.2 pts
 Then torsi on against O is zero, so that
 N12 l − N13 l + f 12 h h1 + f 13 h h1 = 0                       (3) 0.2 pts

Part two : Rear cylinder

                                              N2
          f2

                                              f21h

                    N21
                                Mg
                                                                      0.25 pts

From balance condition in rear wheel :
f 21h − f 2 + Mg sin θ = Ma                                     (4)        0.15 pts
N 2 − N 21 − Mg cos θ = 0                                      (5)         0.15 pts

For pure rolling:

                                              a2
                            f 2 R = Iα2 = I
                                              R
                                     I
                          or f 2 =      a                            (6)
                                     R2
For rolling with sliding:

                    F2 = uk N2                                 (7)

                                                                           0.2 pts

Part Three : Front Cylinder:

                                                     III - 3

                                       N3
           f3

                                       f31h

                     N31
                             Mg
                                                                0.25 pts

From balance condition in the front whee l :
f 31h − f 3 + Mg sin θ = Ma                                                 (8)   0.15 pts
N 3 − N 31 − Mg cos θ = 0                                                  (9)    0.15 pts

For pure rolling:

                    a3
f 3 R = Iα3 = I
                    R
            I
or f 3 =      a                                                                      (10)
           R2
For rolling with sliding:

                    F3 = uk N3                                                       (11)

                                                                                0.2 pts
3. From equation (2), (5) and (9) we get

    m1 gcosθ = N2 – m2 g cosθ+N3 -m3 g cosθ
    N2 + N3 = (m1 +m2 +m3 )gcosθ=7Mgcosθ                                            (12)

And from equation (3), (5) and (8) we get

(N 3 -Mg cos θ) l – (N 2 -Mg cos θ) l=h1 (f2 +Ma-Mg sinθ+f3 +Ma-Mg sinθ)

(N 3 – N2 )= h1 (f2 +2Ma-2Mg sinθ+f3 )/l

                                  Equations 12 and 13 are given 0.25 pts

CASE ALL CYLINDER IN PURE ROLLING

From equation (4) and (6) we get

                                              III - 4

   f21h = (I/R2 )a +Ma-Mg sinθ                                         (14)    0.2 pts

From equation (8) and (10) we get

   f31h = (I/R2 )a +Ma-Mg sinθ                                         (15)    0.2 pts

Then from eq. (1) , (14) and (15) we get

5Mg sinθ - {(I/R2 )a +Ma-Mg sinθ}-{(I/R2 )a +Ma-Mg sinθ}= m1 a

7 Mg sinθ = (2I/R2 +7M)a
                  7 Mg sin θ   7 Mg sin θ
             a=              =                = 0.833g sin θ            (16) 0.35 pts
                           I       0 . 7 MR 2
                 7M + 2 2 7M + 2
                          R             R2

      7M           h         I
N3 =      g cos θ + 1 [( M + 2 ) × 0.833 g sin θ − Mg sin θ ]
       2            l       R
                   h
    = 3.5Mgcos θ + 1 [( M + 0.7M ) × 0.833 g sin θ − Mg sin θ ]
                    l
                        h
    = 3.5 Mg cosθ + 0.41 1 Mg sin θ
                        l

          7M          h     I
N2 =         g cos θ − 1 [( 2 + M ) × 0.833g sin θ − Mg sin θ]
           2          l R
                         h1                 7 Mg sin θ
    = 3.5g cos θ -          [( 0.7 M + M )             − 2 Mg sin θ]
                         l                 0.7M + 7 M
                       h
    = 3.5g cos θ − 0.41 1 Mg sin θ
                        l
                                                                              0.2 pts
The Conditions for pure rolling:

           f 2 ≤ μs N 2              and f 3 ≤ μs N 3
           I2                              I3                                 0.2 pts
                a ≤ μs N 2           and        a ≤ μs N 3
           R 22                            R 23
The left equation becomes
                                             h
0.7M × 0. 833g sin θ ≤ μ s (3.5Mg cosθ − 0.41 1 Mg sin θ )
                                             l
                3.5μ s
tan θ ≤
                             h1
          0.5831+ 0.41μ s
                              l

                                                III - 5

While the right equation becomes
                                            h
0.7m × 0.833g sin θ ≤ μ s (3.5mg cosθ + 0.41 1 mg sin θ )
                                             l
              3.5μ s
tan θ ≤
                         h
        0.5831− 0.41μ s 1
                          l
                                                                                (17)         0.1 pts

CASE ALL CYLINDER SLIDING

From eq. (4) f21h = Ma +uk N2 – Mgsinθ                        (18)                              0.15 pts
From eq. (8) f31h = Ma +uk N3 – Mgsinθ                        (19)                              0.15 pts
From eq. (18) and 19 :
  5Mg sinθ - (Ma + uk N2 – Mg sinθ)- (Ma + uk N3 – Mg sinθ)=m1 a

        7 Mg sin θ − μk N 2 − μk N 3            μ (N + N3)
   a=                                = g sin θ − k 2                               (20)         0.2 pts
                    7M                             7M

N 3 + N 2 = 7Mg cos θ

From the above two equations we get :
                              a = g sin θ − μk g cos θ                                         0.25 pts

The Conditions for complete sliding: are the opposite of that of pure rolling

         f 2 〉 μs N '2             and f 3 〉 μs N '3
         I2                                I3                                 (21)            0.2 pts
              a 〉 μs N '2          and          a 〉 μs N '3
         R 22                              R 32

Where N2 ’ and N3 ’ is calculated in case all cylinder in pure rolling.                   0.1 pts

Finally weget
               3. 5μ s                                                   3. 5μ s              0.2 pts
tan θ 〉                              and                 tan θ 〉
                        h                                                         h
        0. 5831+ 0.41μ s 1                                         0.5831− 0.41μ s 1
                         l                                                         l
The left inequality finally become decisive.

CASE ONE CYLINDER IN PURE ROLLING AND ANOTHER IN SLIDING
CONDITION

{ For example R3 (front cylinders) pure rolling while R2 (Rear cylinders) sliding}

                                                 III - 6

From equation (4) we get

   F21h = m2 a+ uk N2 -m2 g sinθ                          (22)           0.15 pts

From equation (5) we get

   f31h = m3 a+(I/R2 )a -m3 g sinθ                        (23)           0.15 pts

Then from eq. (1) , (22) and (23) we get

m1 g sinθ - { m2 a+uk N2 -m2 g sinθ}-{m3 a+(I/R2 )a -m3 g sinθ}= m1 a

m1 g sinθ + m2 g sinθ + m3 sinθ- uk N2 = (I/R2 +m3 )a + m2 a + m1 a

5Mg sinθ + Mg sinθ + Mg sinθ- uk N2 = (0.7M +M)a + Ma + 5Ma

     7 Mg sin θ − μk N 2                  μN
a=                       = 0.9091g sin θ − k 2                          (24)        0.2 pts
           7.7M                           7.7 M

             h1             I
N3 − N2 =       ( μk N 2 + 2 a + 2Ma − 2Mg sin θ )
              l            R
             h
N 3 − N 2 = 1 ( μk N 2 + 2.7 M × 0.9091g sin θ − 2.7 μk N 2 / 7.7 − 2 Mg sin θ )
              l
                        h
N 3 − N 2 (1 + 0.65 μk 1 ) = 0.4546Mg sin θ
                         l
N 3 + N 2 = 7Mg cos θ

Therefore we get

      7Mg cos θ − 0.4546Mg sin θ
N2 =
                          h
             2 + 0.65μk 1
                           l                                               (25)      0.3 pts
                   7 Mg cos θ − 0.4546Mg sin θ
N 3 = 7Mg cosθ −
                                        h
                            2 + 0.65μk 1
                                         l
Then we can substitute the results above into equation (16) to get the following result

                      μk N 2                  μ 7 g cos θ − 0.4546 g sin θ
a = 0.9091g sin θ −          = 0.9091g sin θ − k                                    (26)
                      7.7M                    7.7                h
                                                      2 + 0.65μk 1
                                                                  l
                                                                                0.2 pts

                                           III - 7

The Conditions for this partial sliding is:
       f 2 ≤ μs N 2′               and f 3 〉 μs N 3′
           I                            I                                  (27)         0.25 pts
             2
               a ≤ μs N 2′        and 2 a 〉 μs N 3′
          R                           R
          where N ′2 and N 3′ are normal forces for pure rolling condition

4. Assumed that after rolling d meter all cylinder start to sliding until reaching the end of
incline road (total distant is s meter). Assummed that ηmeter is reached in t1 second.

v t1 = v o + at 1 = 0 + a1t 1 = a1 t1
              1         1
d = vo t 1 + a1 t 12 = a1 t12
              2         2
        2d
t1 =
         a1
                                                                             0.5 pts
             2d
v t1 = a1       = 2da1 = 2 d 0.833 g sin θ = 1.666 dg sin θ                                     (28)
             a1

The angular velocity after rolling d meters is same for front and rear cylinders:

         vt 1 1
ωt 1 =       =  1.666 dg sin θ                                                                  (29)
          R R
                                                                             0.5 pts

Then the vehicle sliding untill the end of declining road. Assumed that the time needed
by vehicle to move from d position to the end of the declining road is t2 second.

v t 2 = v t1 + a 2 t 2 = 1.666 dg sin θ + a 2 t 2
                     1
s − d = vt 1 t 2 +     a 2 t 22
                     2
                                                                             (30)        0.4 pts
       − v t 1 + vt21 + 2 a2 ( s − d )
t2 =
                      a2
v t 2 = 1.666 dg sin θ − vt 1 + v t21 + 2a 2 ( s − d )

Inserting vt1 and a2 from the previous results we get the final results.

For the angular velocity, while sliding they receive torsion:

                                                    III - 8

τ = μ k NR
      τ μ k NR                                                                              (31)
α =     =
      I    I
                                                            2
                     1                   μ k NR − vt 1 + v t1 + 2 a 2 ( s − d )
ω t2 = ω t1 + αt 2 =   1 .666 dg sin θ +
                     R                       I                a2
                                                                                0.6 pts

                                              III - 9

III. A Heavy Vehicle Moving on An Inclined Road


[User]
In this problem you will analyze the longitudinal motion of a linear molecule, i.e. the motion along
the molecular axis. The rotational motion and the bending of the molecule are not considered. Each
atom is assumed to be connected to its neighbors by a chemical bond, approximated by a massless
spring which obeys Hooke’s law.
  1. Consider a diatomic molecule AB, where atom A has mass mA , atom B has mass mB , and
     the spring constant of the bond is k. Find the angular frequency of vibrations.
  2. Now consider a triatomic molecule ABA, where the two bonds both have spring constant k.
     Find the possible vibrational angular frequencies and sketch the associated motions.

Solution:
In this problem you will analyze the longitudinal motion of a linear molecule, i.e. the motion along
the molecular axis. The rotational motion and the bending of the molecule are not considered. Each
atom is assumed to be connected to its neighbors by a chemical bond, approximated by a massless
spring which obeys Hooke’s law.
   1. Consider a diatomic molecule AB, where atom A has mass mA , atom B has mass mB , and
      the spring constant of the bond is k. Find the angular frequency of vibrations.
   2. Now consider a triatomic molecule ABA, where the two bonds both have spring constant k.
      Find the possible vibrational angular frequencies and sketch the associated motions.
Solution. This is IPhO 1992, problem 2. (Things were a lot easier back then!) The answers are:
          p
  1. ω = k(mA + mB )/(mA mB ). If you know about reduced mass, you can write this down
     immediately.
   2. There are two vibrational modes with nonzero ω.
        • B stays at rest and the A’s move oppositely. This mode clearly has ω = k/mA .
                                                                                p

        • The A’s have the same velocity while B moves against them. Since the center of mass
          has to stay stationary, vB = −2mA vA /mB . This yields
                                              s
                                                     1      2
                                         ω= k           +        .
                                                    mA mB


[User] Satelliit tiirleb ringikujulisel orbiidil (raadiusega $r=\SI{7000}{\kilo\metre}$) ümber maakera, kusjuures satelliidi orbiit on samas tasapinnas Maa orbiidiga ümber Päikese. Kui suure osa ajast keskmiselt veedab satelliit Maa varjus? Maa läbimõõt on $R=\SI{6378}{\kilo\metre}$. Päikeselt tulevad kiired võib lugeda paralleelseteks ja Maa liikumise ühe satelliidi orbiiditaalperioodi jooksul tühiseks.

Solution:
\begin{center}
	\includegraphics[width=0.9\linewidth]{2011-v2g-02-lah}
\end{center}
Ringikujulisel orbiidil on satelliidi kiirus kogu orbitaalperioodi jooksul konstantne ja seetõttu on varjus veedetud osa ajast võrdne orbiidi varjus oleva osa pikkuse ja kogu orbiidi pikkuse suhtega, mis on ülal toodud jooniselt leitav kui
\[
k=\frac{2 \alpha r}{2 \pi r}=\frac{\arcsin \left(\frac{R}{r}\right)}{\pi}=\SI{36,5}{\%}
\]

Satelliit

Satelliit tiirleb ringikujulisel orbiidil (raadiusega $r=\SI{7000}{\kilo\metre}$) ümber maakera, kusjuures satelliidi orbiit on samas tasapinnas Maa orbiidiga ümber Päikese. Kui suure osa ajast keskmiselt veedab satelliit Maa varjus? Maa läbimõõt on $R=\SI{6378}{\kilo\metre}$. Päikeselt tulevad kiired võib lugeda paralleelseteks ja Maa liikumise ühe satelliidi orbiiditaalperioodi jooksul tühiseks.

Ringikujulisel orbiidil on satelliidi kiirus kogu orbitaalperioodi jooksul konstantne ja seetõttu on varjus veedetud osa ajast võrdne orbiidi varjus oleva osa pikkuse ja kogu orbiidi pikkuse suhtega.


[User]
Plasmonic Steam Generator                                                           T2

Introduction
In this problem we study an efficient process of steam production that has been demonstrated to
work experimentally. An aqueous solution of spherical nanometer-sized silver spheres
(nanoparticles) with only about       particles per liter is illuminated by a focused light beam. A
fraction of the light is absorbed by the nanoparticles, which are heated up and generate steam
locally around them without heating up the entire water solution. The steam is released from the
system in the form of escaping steam bubbles. Not all details of the process are well understood at
present, but the core process is known to be absorption of light through the so-called collective
electron oscillations of the metallic nanoparticles. The device is known as a plasmonic steam
generator.

Figure 2.1 (a) A spherical charge-neutral nanoparticle of radius R placed at the center of the coordinate
system. (b) A sphere with a positive homogeneous charge density (red), and containing a smaller spherical
charge-neutral region (0, yellow) of radius , with its center displaced by              . (c) The sphere with
positive charge density of the nanoparticle silver ions is fixed in the center of the coordinate system. The
center of the spherical region with negative spherical charge density – (blue) of the electron cloud is
displaced by     , where          . (d) An external homogeneous electric field                    . For time-
dependent     , the electron cloud moves with velocity               . (e) The rectangular vessel (         )
containing the aqueous solution of nanoparticles illuminated by monochromatic light propagating along the
 -axis with angular frequency      and intensity .

A single spherical silver nanoparticle
Throughout this problem we consider a spherical silver nanoparticle of radius                      and
with its center fixed at the origin of the coordinate system, see Fig. 2.1(a). All motions, forces and
driving fields are parallel to the horizontal -axis (with unit vector ). The nanoparticle contains
free (conduction) electrons moving within the whole nanoparticle volume without being bound to
any silver atom. Each silver atom is a positive ion that has donated one such free electron.
    Find the following quantities: The volume and mass of the nanoparticle, the
2.1 number and charge density of silver ions in the particle, and for the free electrons                0.7
    their concentration , their total charge , and their total mass .

                                                Page 1 of 4

                           Plasmonic Steam Generator                                                       T2

The electric field in a charge-neutral region inside a charged sphere
For the rest of the problem assume that the relative dielectric permittivity of all materials is       .
Inside a charged sphere of homogeneous charge density and radius R is created a small spherical
charge-neutral region of radius     by adding the opposite charge density      , with its center displa-
ced by             from the center of the R-sphere, see Fig. 2.1(b).

       Show that the electric field inside the charge-neutral region is homogenous of the form
2.2                                                                                1.2
             (     ) , and determine the pre-factor .

The restoring force on the displaced electron cloud
In the following, we study the collective motion of the free electrons, and therefore model them as a
single negatively charged sphere of homogeneous charge density               with a center position ,
which can move along the x-axis relative to the center of the positively charged sphere (silver ions)
fixed at the origin of the coordinate system, see Fig. 2.1(c). Assume that an external force
displaces the electron cloud to a new equilibrium position                   with          . Except for
tiny net charges at opposite ends of the nanoparticle, most of its interior remains charge-neutral.

       Express in terms of    and n the following two quantities: The restoring force exerted
2.3                                                                                1.0
       on the electron cloud and the work     done on the electron cloud during displacement.

The spherical silver nanoparticle in an external constant electric field
A nanoparticle is placed in vacuum and influenced by an external force       due to an applied static
homogeneous electric field              , which displaces the electron cloud the small distance     ,
where           .

    Find the displacement     of the electron cloud in terms of       , and determine the
2.4 amount         of electron charge displaced through the yz-plane at the center of the 0.6
    nanoparticle in terms of            .

The equivalent capacitance and inductance of the silver nanoparticle
For both a constant and a time-dependent field , the nanoparticle can be modeled as an equivalent
electric circuit. The equivalent capacitance can be found by relating the work     , done on the
separation of charges     , to the energy of a capacitor, carrying charge . The charge separation
will cause a certain equivalent voltage across the equivalent capacitor.

2.5a Express the systems equivalent capacitance C in terms of       and , and find its value.      0.7
       For this capacitance, determine in terms of       and     the equivalent voltage      that
2.5b                                                                                0.4
       should be connected to the equivalent capacitor in order to accumulate the charge      .

                                              Page 2 of 4

                           Plasmonic Steam Generator                                                        T2

For a time-dependent field , the electron cloud moves with velocity           , Fig. 2.1(d). It has
the kinetic energy       and forms an electric current flowing through the fixed yz-plane. The
kinetic energy of the electron cloud can be attributed to the energy of an equivalent inductor of
inductance carrying the current .

2.6a Express both        and    in terms of the velocity .                                          0.7
       Express the equivalent inductance in terms of particle radius , the electron charge
2.6b                                                                                0.5
       and mass     , the electron concentration , and calculate its value.

The plasmon resonance of the silver nanoparticle
From the above analysis it follows that the motion, arising from displacing the electron cloud from
its equilibrium position and then releasing it, can be modeled by an ideal LC-circuit oscillating at
resonance. This dynamical mode of the electron cloud is known as the plasmon resonance, which
oscillates at the so-called angular plasmon frequency .

       Find an expression for the angular plasmon frequency       of the electron cloud in terms
2.7a                                                                                0.5
       of the electron charge and mass      , the electron density , and the permittivity .
       Calculate    in rad/s and the wavelength       in nm of light in vacuum having angular
2.7b                                                                                0.4
       frequency         .

The silver nanoparticle illuminated with light at the plasmon frequency
In the rest of the problem, the nanoparticle is illuminated by monochromatic light at the angular
plasmon frequency         with the incident intensity                               . As the wavelength
is large,         , the nanoparticle can be considered as being placed in a homogeneous harmonical-
ly oscillating field                 (    ) . Driven by , the center ( ) of the electron cloud
oscillates at the same frequency with velocity                     and constant amplitude          . This
oscillating electron motion leads to absorption of light. The energy captured by the particle is either
converted into Joule heating inside the particle or re-emitted by the particle as scattered light.
    Joule heating is caused by random inelastic collisions, where any given free electron once in a
while hits a silver ion and loses its total kinetic energy, which is converted into vibrations of the
silver ions (heat). The average time between the collisions is                      , where for silver
nanoparticle we use                         .

     Find an expression for the time-averaged Joule heating power      in the nanoparticle
2.8a as well as the time-averaged current squared 〈 〉, which includes explicitly the time- 1.0
     averaged velocity squared 〈 〉 of the electron cloud.
     Find an expression for the equivalent ohmic resistance            in an equivalent resistor-
2.8b model of the nanoparticle having the Joule heating power           due to the electron cloud 1.0
     current I. Calculate the numerical value of     .

                                              Page 3 of 4

                           Plasmonic Steam Generator                                                      T2

The incident light beam loses some time-averaged power           by scattering on the oscillating elec-
tron cloud (re-emission).       depends on the scattering source amplitude , charge , angular fre-
quency      and properties of the light (the speed of light and permittivity in vacuum). In terms
of these four variables,      is given by                     .

        By use of     , find an expression of the equivalent scattering resistance      (in ana-
2.9                                                                                1.0
        logy with     ) in an equivalent resistor-model, and calculate its value.

The above equivalent circuit elements are combined into an LCR series circuit model of the silver
nanoparticle, which is driven by a harmonically oscillating equivalent voltage            (     )
determined by the electric field of the incident light.

      Derive expressions for the time-averaged power losses         and     involving the
2.10a amplitude     of the electric field in the incident light at the plasmon resonance 1.2
             .
2.10b Calculate the numerical value of      ,       , and         .                                0.3

Steam generation by light
An aqueous solution of silver nanoparticles is prepared with a concentration                         .
It is placed inside a rectangular transparent vessel of size                                      and
illuminated by light at the plasmon frequency with the same intensity                       at normal
incidence as above, see Fig. 2.1(e). The temperature of the water is               and we assume, in
fair agreement with observations, that in steady state all Joule heating of the nanoparticle goes to
the production of steam of temperature               , without raising the temperature of the water.

The thermodynamic efficiency of the plasmonic steam generator is defined by the power ratio
              , where     is the power going into the production of steam in the entire vessel, while
    is the total power of the incoming light that enters the vessel.

Most of the time any given nanoparticle is surrounded by steam instead of water, and it can thus be
described as being in vacuum.

         Calculate the total mass per second       of steam produced by the plasmonic steam ge-
2.11a                                                                                0.6
         nerator during illumination by light at the plasmon frequency and intensity .
         Calculate the numerical value of the thermodynamic efficiency        of the plasmonic
2.11b                                                                                0.2
         steam generator.

                                                Page 4 of 4

Solution:
Plasmonic Steam Generator                                                          T2

                                                 Solutions

A single spherical silver nanoparticle
      Volume of the nanoparticle:                                          .
      Mass of the nanoparticle:                                        .
      Number of ions in the nanoparticle:                                      .
2.1 Charge density                                      , charge density           .                     0.7

      Electrons’ concentration                                 .
      Total charge of free electrons                               .
      Total mass of free electrons                                         .

The electric field in a charge-neutral region inside a charged sphere
      For a sphere with radius and constant charge density , for any point inside the sphere
      designated by radius-vector                 ) Gauss's law yields directly
                , where     is the unit radial vector pointing away from the center of the
      sphere. Thus,              .
    Likewise, inside another sphere of radius    and charge density       the field is
2.2       , where is the radius-vector of the point in the coordinate system with the origin 1.2
    in the center of this sphere.
    Superposition of the two charge configurations gives the setup we want with
       . So
     inside the charge-free region |         |        the field is
                      or             with pre-factor

The restoring force on the displaced electron cloud
      With               and             we have from above that approximately the field induced
      inside the particle is                . The number of electrons on the particle’s border that
      produced         is negligibly smaller than the number of electrons inside the particle, so
2.3                                                        (note the antiparallel attractive force       1.0
      is proportional to the displacement that it is similar to Hooke’s law).
      The     work        done     on    the       electron      cloud    to           shift   it   is
               ∫                     (             )

The spherical silver nanoparticle in an external constant electric field
    Inside the metallic particle in the steady state the electric field must be equal to 0. The
2.4 induced field (from 2.2 or 2.3) compensates the external field:                         , so 0.6

                                                 Page 1 of 3

                             Plasmonic Steam Generator                                                          T2

                             .
       Charge displaced through the       -plane is the total charge of electrons in the cylinder of
       radius and height :                                           .

The equivalent capacitance and inductance of the silver nanoparticle
       The electric energy       of a capacitor with capacitance        holding charges         is
2.5a              . The energy of such capacitor is equal to the work (see 2.3) done to separate          0.7
       the charges (see 2.4), thus                                            .

       Equivalent scheme for a capacitor reads:                   . Combining charge from (2.4) and
2.5b                                                                                0.4
       capacitance from (2.5a) gives                          .

     The kinetic energy of the electron cloud is defined as the kinetic energy of one electron
     multiplied      by      the     number       of      electrons       in     the     cloud
2.6a                             (        ).                                                   0.7
     The current is the charge of electrons in the cylinder of area            and height
     divided by time    (or simply the time derivative of charge       ), thus               .
     The energy carried by current in the equivalent circuit with inductance is
2.6b is, in fact, the kinetic energy of electrons    . Taking the energy and current from 0.5
     (2.6a) gives                                 .

The plasmon resonance of the silver nanoparticle
     From the LC-circuit analogy we can directly derive                        √             .
2.7a Alternatively it is possible to use the harmonic law of motion in (2.3) and get the same 0.5
     result for the frequency.
                             rad/s, for light with angular frequency                the wavelength is
2.7b                                                                                0.4
                                   .

The silver nanoparticle illuminated with light at the plasmon frequency
       The velocity of an electron                                                . The time-averaged

       kinetic energy on the electron 〈      〉        〈   〉        〈   〉. During time     each electron
       hits an ion one time. So the energy lost in the whole nanoparticle during time                is
2.8a                                                                                1.0
                   〈    〉             〈    〉. Time-averaged Joule heating power

                                 〈   〉(          ).
       The expression for current is taken from (2.6a), squared and averaged

                                                  Page 2 of 3

                              Plasmonic Steam Generator                                                               T2

     〈 〉                  〈   〉   ( ) 〈       〉.

     The average time between the collisions is            , so each electron oscillates many
     times before it collides with an ion. The oscillating current
2.8b                   produces the heat in the resistance        equal to                〈 〉, 1.0
     that together with results from (2.8a) leads to         〈 〉
                                                                                .

     For equivalent scattering resistance                  〈 〉
                                                                   and for harmonic oscillations we can
2.9 average the velocity squared over one period of oscillations, so 〈                        〉            . 1.0
     Together it yields                        〈   〉
                                                                            .

      Ohm’s law for a LCR serious circuit is                                            . At the resonance
                                                           √                    (   )

      frequency time-averaged voltage squared is 〈                   〉      〈 〉                         〈 〉.

2.10a And from (2.5b) 〈       〉                        , so Ohm’s law results in 〈 〉                       .
                                                                                1.2
      The time-averaged power losses are                           〈 〉                            and

                                          〈         〉.

2.10b Starting with the electric field amplitude               √                            , we calculate 0.3
                        and                   .

Steam generation by light
      Total number of nanoparticles in the vessel:                               Then the
      total time-averaged Joule heating power:                         kW. This power goes
2.11a into the steam generation:              , with                                                           0.6
                                           . Thus the mass of steam produced in one second
      is:                                .

      The power of light incident on the vessel
2.11b          , and the power directed for steam production by nanoparticles is given in                      0.2
      2.11a. Efficiency of the process is                  = 0.498.

     Total                                                                                12.0

                                                   Page 3 of 3

Introduction