[User]
Question A4
A positive point charge q is located inside a neutral hollow spherical conducting shell. The shell
has inner radius a and outer radius b; b − a is not negligible. The shell is centered on the origin.

                                                  y

                                                  b

                                                         a

                                                                      x

  a. Assume that the point charge q is located at the origin in the very center of the shell.

        i. Determine the magnitude of the electric field outside the conducting shell at x = b.
       ii. Sketch a graph for the magnitude of the electric field along the x axis on the answer
           sheet provided.
       iii. Determine the electric potential at x = a.
       iv. Sketch a graph for the electric potential along the x axis on the answer sheet provided.

  b. Assume that the point charge q is now located on the x axis at a point x = 2a/3.

        i. Determine the magnitude of the electric field outside the conducting shell at x = b.
       ii. Sketch a graph for the magnitude of the electric field along the x axis on the answer
           sheet provided.
       iii. Determine the electric potential at x = a.
       iv. Sketch a graph for the electric potential along the x axis on the answer sheet provided.
       v. Sketch a figure showing the electric field lines (if any) inside, within, and outside the
          conducting shell on the answer sheet provided. You should show at least eight field lines
          in any distinct region that has a non-zero field.

                   Copyright c 2014 American Association of Physics Teachers

2014 USA Physics Olympiad Exam          Part A                             6

   STOP: Do Not Continue to Part B

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

               Copyright c 2014 American Association of Physics Teachers

2014 USA Physics Olympiad Exam                Part B                                               7

Part B

Solution:
Question A4
A positive point charge q is located inside a neutral hollow spherical conducting shell. The shell
has inner radius a and outer radius b; b − a is not negligible. The shell is centered on the origin.

                                                   y

                                                    b

                                                         a

                                                                          x

  a. Assume that the point charge q is located at the origin in the very center of the shell.

        i. Determine the magnitude of the electric field outside the conducting shell at x = b.

                                                   Solution
          We apply Gauss’s law for a sphere with radius r > b centered about the origin. Since
          the shell is neutral, the enclosed charge is q, so by spherical symmetry
                                                               q
                                                 E(r) =
                                                             4π0 r2

          outside the shell. Just outside the shell, the field is q/4π0 b2 .

       ii. Sketch a graph for the magnitude of the electric field along the x axis on the answer
           sheet provided.

                                                   Solution
          Since the shell is conducting, the electrostatic field is zero inside it. By Gauss’s law, this
          is achieved by having a charge of −q on the inner surface r = a and a charge of q on the
          outer surface r = b, both uniformly distributed.
          For r < a, we can apply Gauss’s law again to conclude E(r) = 4πq0 r2 , just as it is outside
          the shell.

                   Copyright c 2014 American Association of Physics Teachers

2014 USA Physics Olympiad Exam               Part A                                              12

     iii. Determine the electric potential at x = a.

                                                Solution
         The shell is a conductor, so it is an equipotential surface. Then the potential at r = a is
         same as the potential at r = b. However, outside the shell the field looks just like that
         of a point charge q at the origin, so
                                                              q
                                          V (a) = V (b) =          .
                                                            4π0 b

      iv. Sketch a graph for the electric potential along the x axis on the answer sheet provided.

                                                Solution
         As we’ve shown above, the potential is proportional to 1/r outside r = b, and is constant
         between r = a and r = b. Then the potential for r < a is not proportional to 1/r.
         Instead, it is a 1/r curve shifted by a constant, so that the potential is continuous at
         r = a.

                 Copyright c 2014 American Association of Physics Teachers

2014 USA Physics Olympiad Exam                 Part A                                             13

  b. Assume that the point charge q is now located on the x axis at a point x = 2a/3.

       i. Determine the magnitude of the electric field outside the conducting shell at x = b.

                                                  Solution
         The conducting shell acts like a Faraday cage. As in the previous part, by Gauss’s law,
         we must have a charge of −q on the inner surface, so a charge of q on the outer surface.
         The charge on the inner surface is distributed non-uniformly to perfectly cancel out
         the asymmetric field of the point charge; these two contributions sum to exactly zero
         everywhere outside r = a. Then by spherical symmetry, the charges on the outer surface
         are uniformly distributed.
         One might wonder why the charges on the outer and inner surfaces can’t both be non-
         uniformly distributed. A more rigorous argument would appeal to the uniqueness the-
         orems for electrostatics: given the setup we’ve given, there is only one way to satisfy
         all the boundary conditions, so the configuration we gave above must be it. The gen-
         eral principle in that in electrostatics, the only information that can be seen across a
         shielding conducting shell is the total charge.
         In any case, by the same logic as in part (a),
                                                           q
                                                E(r) =
                                                         4π0 r2

         outside the shell. Just outside the shell, the field is q/4π0 b2 .

      ii. Sketch a graph for the magnitude of the electric field along the x axis on the answer
          sheet provided.

                                                  Solution
         For r < b, the field is just that of a point charge at the origin, by the shell theorem. The
         field inside is more complicated because it depends on the distribution of charge on the
         inner surface; all that is required is that it diverges at x = 2a/3 and is higher at x = a
         than x = −a.

                 Copyright c 2014 American Association of Physics Teachers

2014 USA Physics Olympiad Exam               Part A                                            14

         Incidentally, one can find the field for r < a exactly using the method of image charges:
         for r < a, the shielding charges on the inner surface produce the exact same field as a
         single point charge would. The location of this “image” charge can be found by inverting
         the original point charge about the circle r = a.

     iii. Determine the electric potential at x = a.

                                               Solution
         By the same logic as in part (a),
                                                          q
                                              V (a) =          .
                                                        4π0 b

      iv. Sketch a graph for the electric potential along the x axis on the answer sheet provided.

                                               Solution
         Again, the potential is proportional to 1/r outside r = b, and is constant between r = a
         and r = b. The potential inside is more complicated, diverging at x = 2a/3.

      v. Sketch a figure showing the electric field lines (if any) inside, within, and outside the
         conducting shell on the answer sheet provided. You should show at least eight field lines
         in any distinct region that has a non-zero field.

                                               Solution
         The field should be spherically symmetric outside the shell, zero within the shell, and
         nonuniform inside. The field lines should terminate perpendicular to the conductor.

                 Copyright c 2014 American Association of Physics Teachers

2014 USA Physics Olympiad Exam          Part A                             15

               Copyright c 2014 American Association of Physics Teachers

2014 USA Physics Olympiad Exam          Part A                             16

   STOP: Do Not Continue to Part B

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

               Copyright c 2014 American Association of Physics Teachers

2014 USA Physics Olympiad Exam                   Part B                                              17

Part B

USAPhO 2014 Question A4


[User]
Show that any surface of charge density σ with electric fields E1 and E2 immediately
    on its two sides experiences a force σ(E1 + E2 )/2 per unit area. (This is a generalization of the
    example above, where one side was inside a conductor.)

Solution:
Let E be the field due to all the other charges. Again, we have E1 = E + σ/2ϵ0 n̂ and
    E2 = E − σ/2ϵ0 n̂. Thus, E = 12 (E1 + E2 ), and the force per area is σE.

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


[User]
1. GRAVITATIONAL RACING (11 points)

—
Maté Vigh and Jaan Kalda. While in generic
case, the dynamics of three gravitationally inter-
acting bodies is complicated and chaotic, there
are special cases when the dynamics is regular.
In particular, there are cases when bodies per-
formperiodicmotion. Thesimplestcaseofsuch
periodic motion is when all three bodies, posi-
tioned at the vertices of an equilateral triangle,
rotate as a rigid body. In what follows we con-
sider a more complicated periodic motion.
Relatively recently 1, it was discovered that
three equal point masses can move periodically
along a common 8-shaped trajectory shown in
the figure (arrow denotes the direction of mo-
tion). This figure is based on a computer simula-
tion and has a correct shape. If needed, you can
measure distances from the enlarged version of
the figure (on a separate sheet) using a ruler.
Letusenumeratethethreebodieswithnum-
bers 1, 2, and 3, according to the order in which
they pass the leftmost point P shown in the fig-
ure. Let O2 and O3 denote the positions of
the bodies 2 and 3, respectively at that moment
when the body 1 is passing the middle point O.
Similarly, let P2 and P3 denote the positions of
the bodies 2 and 3, respectively at that moment
when the body 1 is passing the leftmost point P.
Let T denote the full period of motion of each
of the bodies along this 8-shaped trajectory.
i)(2points)Expressthefollowingtraveltimesfor
one of the bodies: (a) from O2 to O; (b) from
O3 to P2.
ii) (1 point) Let ⃗ v1, ⃗ v2, and ⃗ v3 denote the ve-
locities of the three bodies at a certain moment
of time. Write down an equality relating these
three vectors to each other.
iii) (2 points) Prove that the total angular mo-
mentum of this system is zero.
iv) (2 points) Construct the positions of the
points O2 and O3 (use the figure on the separate
sheet). Motivate your construction.
v)(2points)Constructthepositionsofthepoints
P2 and P3 (use the figure on the separate sheet).
Find two independent ways of construction and
motivate your methods.
vi) (2 points) Find the ratio of the speed of the
bodies at point O to the speed at point P.

Solution:
1. GRAVITATIONAL RACING

i) (a)Sinceallthreebodies movealongthesame
trajectory, they must be T
3 away from each other
at any moment of time. Thus, it takes T
3 to get
from O2 to O.
(b) From symmetry, time taken to go from
P to O must be T
4 . Furthermore, it takes T
3 to
get from P2 to P and from O to O3. This means
that it takes T
3 + T
3 + T
4 = 11T
12 to get from P2 to
O3 or T − 11T
12 = T
12 to get from O3 to P2.
ii) Since there are no external forces at play, the
centre of mass of the three body system must
stay in place and, due to symmetry, be located
at O. Thus,⃗ r1+⃗ r2+⃗ r3 =0, where⃗ r1,⃗ r2 and⃗ r3
are position vectors from O. Differentiating,
⃗ v1 +⃗ v2 +⃗ v3 =0. (1)
iii) The total angular momentum is conserved.
Thus, we can find the angular momentum at a
moment of time that’s most convenient for us,
suchastheconfigurationwhenoneofthebodies
is at O. Due to symmetry,⃗ r2 =−⃗ r3 and⃗ v2 =⃗ v3.
The total angular momentum is then
J = m⃗ r1 ×⃗ v1 +m⃗ r2 ×⃗ v2 +m⃗ r3 ×⃗ v3 =
= m(⃗ r2 ×⃗ v2 +⃗ r3 ×⃗ v3)= m(⃗ r2 ×⃗ v2 −⃗ r2 ×⃗ v2)=0.
iv) Using equation ?? and the fact that⃗ v2 =⃗ v3,
⃗ v1 +2⃗ v2 = 0 must hold. Thus, ⃗ v2 = 1
2⃗ v1. Note
that this can only happen when ⃗ v1, ⃗ v2 and ⃗ v3
are parallel. Since the velocities of the bodies
are tangent to the trajectory at their locations,
thetrajectoriesofthebodiesmusthavethesame
slope at O, O2 and O3. This gives a convenient
way of reconstructing the positions of O2 and
O3 using a ruler as shown in the figure in red.
v) We can use symmetry to deduce that r2,x =
r3,x and r2,y =−r3,y. Additionally, r1,x +r2,x +
r3,x = 0, so r2,x = r3,x = −1
2 r1,x. This gives a
simple way to find the x-coordinates of P2 and
P3 andreconstructtheirpositionsshowninblue
in the figure.
Let’s apply the conservation of angular mo-
mentum at point P. Due to symmetry, the ve-
locity vectors at P2 and P3 must intersect at the
x-axis. For the total angular momentum to be 0,
they must intersect at P. Therefore, P2 and P3
have tangents that pass through P. P2 and P3
can then be reconstructed as shown in green in
the figure.
O3
O2
P2
P3
d
ro Q Rp
α
vi) The general idea is to apply energy conser-
vation in both configurations and use the radius
of curvature at P to find the value of Gm. From
the figure, Rp = 41.7, r0 = 145.2, d = 242 and
α=11.6deg (the units for distance are arbitrary,
only the ratios carry any meaning).
The total energy at O is
E=
mv2
1,o
2
+
mv2
2,o
2
+
mv2
3,o
2
−
Gm2
r12,o
−
Gm2
r23,o
−
Gm2
r31,o
.
We have v1,o = 2v2,o = 2v3,o = vo and r12,o =
r31,o = 1
2 r23,o = ro. Thus,
E =
3mv2
o
4
−
5Gm2
2ro
. (2)
Total Energy at P is
E=
mv2
1,p
2
+
mv2
2,p
2
+
mv2
3,p
2
−
Gm2
r12,p
−
Gm2
r23,p
−
Gm2
r31,p
.
Additionally, r12,p = r31,p = d and r23,p =
2dsinα. Furthermore, applying equation ??
on the y-axis, v1,p − v2,p sinα− v3,p sinα = 0.
Thus, v1,p = vp =2v2,p sinα=2v3,p sinα since
v2,p = v3,p. The total energy at P is then
E =
mv2
p
2
(
1+
1
2sin2
α
)
−
Gm2
d
(
2+
1
2sinα
)
=
=6.68mv2
p −4.49
Gm2
d
.
(3)
When a body is at P, the gravitational force is
equal to the centrifugal force. This means that
mv2
p
Rp
=2
Gm2
d2
cosα=1.96
Gm2
d2
,
Gm=0.510
v2
pd2
Rp
. (4)
Combining equations ??, ?? and ?? gives
3v2
o
4
=6.68v2
p +0.510
v2
pd2
Rp
(
5
2ro
−
4.49
d
)
,
rearranging,
vo
vp
=
√
4
3
(
6.68+0.510
d2
Rp
(
5
2r0
−
4.49
d
))
=2.8.

GRAVITATIONAL RACING


[User]
\begin{wrapfigure}{r}{0.3\textwidth}
	\vspace{-25pt}
	\begin{center}
		\includegraphics[width=0.32\textwidth]{2017-v2g-09-laengudjoonis}
	\end{center}
	\vspace{-10pt}
\end{wrapfigure}
Two balls, each with a charge $Q$, are firmly fixed so that the distance between their centers is $2d$. A third ball with a charge $-q$ and mass $m$ is placed exactly in the middle of the two balls. The third ball can only move along the $y$-axis (see figure). Find the period $T$ of the third ball’s small oscillations along the $y$-axis.

Solution:
Olgu keskmise kera nihe $y$. Keskmise kera kaugus kummastki äärmisest kerast on nüüd $l=\sqrt{d^2+y^2}$. Kumbki äärmine kera mõjutab keskmist kera jõuga $F=\frac{kqQ}{l^2}$, mille vertikaalkomponent on $F_y=-\frac{y}{l}F$. Kokku mõjub keskmisele kerale vertikaaljõud $2F_y$, millest saame seose keskmise kera kiirenduse jaoks $ma=2F_y$. Et $y\ll d$, siis $l\approx d$, millest saame lihtsustada liikumisvõrrandi kujule
\[
a=-\frac{2kQq}{md^3}y.
\]
See võrrand kirjeldab harmoonilist võnkumist perioodiga
\[ T = 2\pi \sqrt{\frac{md^3}{2kQq}}.\]
{\em Märkus.} perioodi ja sageduse valem on võimalik leida analoogiast vedru võnkumisega, kus $ma = -kx$ ja $T=2\pi \sqrt{m/k}$.

Laengud

\begin{wrapfigure}{r}{0.3\textwidth}
	\vspace{-25pt}
	\begin{center}
		\includegraphics[width=0.32\textwidth]{2017-v2g-09-laengudjoonis.pdf}
	\end{center}
	\vspace{-10pt}
\end{wrapfigure}

Kaks kera, kumbki laenguga $Q$, on liikumatult fikseeritud nii, et nende keskpunktide kaugus on $2d$. Täpselt nende kerade keskele paigutatakse kolmas kera laenguga $-q$ ning massiga $m$, mis saab liikuda ainult mööda $y$-telge (vt joonis). Leidke selle kolmanda kera väikeste $y$-suunaliste võnkumiste periood $T$.

Väikeste $y$-suunaliste võnkumiste jaoks on kasulik vaadata, missugune jõud kerale mõjub, kui seda väikse distantsi $y$ võrra tasakaaluasendist eemale nihutada.


[User]
Klaaskuubi neli tahku on värvitud mustaks nõnda,
et värvimata jäänud tahud paiknevad kõrvuti (omavad ühist serva). Missugune peab olema klaasi murdumisnäitaja $n$, et ka värvimata tahud
paistaksid mustadena?

Solution:
Mustaks värvitud tahkudelt valgus peegelduda ei saa, toimub neeldumine. Ülesande tingimus on täidetud, kui värvimata tahust kuupi sisenev valgus ei saa
väljuda läbi kõrvaltahu (toimub sisepeegeldus). Sisepeegelduse kriitilise nurga saame murdumisseadusest:
\[
\sin \gamma_{C}=\frac{1}{n}.
\]

\begin{center}
	\includegraphics[width=0.45\linewidth]{2009-v2g-07-lah}
\end{center}

Kiir siseneb kuupi, kui $\alpha<\gamma_{C}$. Langemisnurk kõrvaltahule on $\beta = \ang{90} - \alpha$. Kiir väljub kuubist, kui
\[
\beta<\gamma_{C}, \quad \text { ehk } \quad \ang{90}<\gamma_{C}+\alpha<2 \gamma_{C}.
\]
Et kiir ei saaks kuubist väljuda, peab kehtima:
\[
\gamma_{C}<\ang{45}, \quad \text { ehk } \quad n>1 / \sin \ang{45}=\sqrt{2}.
\]

Klaaskuup

Klaaskuubi neli tahku on värvitud mustaks nõnda,
et värvimata jäänud tahud paiknevad kõrvuti (omavad ühist serva). Missugune peab olema klaasi murdumisnäitaja $n$, et ka värvimata tahud
paistaksid mustadena?


[User]
Consider two adjacent blackbodies with similar temperatures, exchanging energy by
    radiation. Show that the rate of change of the temperature difference is approximately proportional
    to the temperature difference itself. This is an example of Newton’s law of cooling, which implies

    the temperature difference decays exponentially. How does the time scale for cooling depend on the
    average temperature of the blackbodies?
[3] Problem 18 (IPhO 1992). A satellite is a sphere of diameter D orbiting about the Earth.

     (a) First suppose the satellite is perfectly black. Ignoring the effect of the Earth, find its temper-
         ature T in terms of the temperature T⊙ of the Sun, the radius R⊙ of the Sun, and the radius
         R of Earth’s orbit.

     (b) How does the answer change if the satellite has a uniform emissivity e(f ) = e0 < 1?

     (c) If an engineer wishes to make T as small as possible using a special paint, which can have an
         arbitrary e(f ), sketch how the function e(f ) should look.

[3] Problem 19 (IPhO 1996). Two perfectly black surfaces of temperatures Th and Tℓ are parallel to
    each other in vacuum, and the net heat flux from the hotter surface to the colder one is P .

    Now suppose that N parallel, thermally insulating, perfectly black plates are placed in between
    them. This shielding reduces the heat flux to P ′ . Find P ′ in terms of P .

       Example 10

       In problem 19 you considered some ideally black surfaces, but things get more complicated
       when the surfaces have general emissivity. For simplicity, consider two walls facing each other,
       with emissivities e1 and e2 . Verify that when both walls have temperature T , there is no net
       energy flow from one to the other.

       Solution

       Let I0 = σT 4 . The second wall emits radiation of intensity e2 I0 , so in equilibrium it must
       absorb this amount as well. Verifying this takes a little work.

       The first wall emits radiation of intensity e1 I0 , and a fraction e2 of it gets absorbed by the
       second wall. The rest bounces back to the first wall, and if it doesn’t get absorbed by the first
       wall, then it comes back to the second wall again for another chance to get absorbed. The
       total intensity transferred from the first wall to the second is the sum of an infinite series,
                                                                                e1 e2 I0
                      I1→2 = (e1 I0 ) (e2 + (1 − e2 )(1 − e1 )e2 + . . .) =                   .
                                                                              e1 + e2 − e1 e2
       This isn’t the expected result, because there’s another contribution. When the second wall
       emits radiation, it can reflect off the first wall and then get reabsorbed by the second wall.
       So the second wall absorbs a total radiation intensity due to its own emission, of

                                                                                (1 − e1 )e22 I0
              I2→2 = (e2 I0 ) ((1 − e1 )e2 + (1 − e1 )(1 − e2 )(1 − e1 )e2 + . . .) =                    .
                                                                                e1 + e2 − e1 e2

       When you sum these contributions, you get e2 I0 as expected. This is a typical example of
       the power of thermodynamics: it is often easy to tell what the answer has to be, but the way
       that answer comes about can be a bit complicated.

       Example 11

       Suppose a black sphere of radius r is placed inside a large container of arbitrary shape, whose
       walls are a blackbody with temperature T . In the steady state, what is the rate of power
       absorbed by the sphere from the walls?

       Solution
       Radiation is continually emitted from all points on the walls in all directions, so it would
       seem that we have to do a complicated integral to find how much hits the sphere, which
       would be hard even if we were given the container’s shape. But the problem is actually
       trivial due to the second law of thermodynamics!

       Let’s suppose the sphere itself is also at temperature T . Then it sends power (4πr2 )σT 4
       to the walls, since all radiation that exits the sphere hits the walls. But if the sphere and
       walls are at the same temperature, a temperature difference cannot spontaneously appear.
       Therefore, the power going from the walls to the sphere must also be (4πr2 )σT 4 . And
       since this quantity can also be expressed as a geometric integral, it can’t depend on the
       temperature of the sphere, so it must be true for any sphere temperature.

       For a tough problem which uses a variant of this idea, see EFPhO 2010, problem 2. An
       improved solution due to Stefan Ivanov is available here.

Solution:
By the Stefan–Boltzmann law, the net energy transfer rate is proportional to T14 − T04 .
    Letting ∆T = T1 − T0 ≪ T0 , we have

                               T14 − T04 = T04 ((1 + (∆T )/T0 )4 − 1) ≈ 4T03 ∆T

    by the binomial theorem. This is proportional to ∆T as desired. Therefore, d∆T /dt = −k∆T for
    a constant k, so ∆T ∼ e−kt , as desired, where k ∝ T 3 .
[3] Problem 18 (IPhO 1992). A satellite is a sphere of diameter D orbiting about the Earth.

      (a) First suppose the satellite is perfectly black. Ignoring the effect of the Earth, find its temper-
          ature T in terms of the temperature T⊙ of the Sun, the radius R⊙ of the Sun, and the radius
          R of Earth’s orbit.

     (b) How does the answer change if the satellite has a uniform emissivity e(f ) = e0 < 1?

      (c) If an engineer wishes to make T as small as possible using a special paint, which can have an
          arbitrary e(f ), sketch how the function e(f ) should look.

    Solution.    (a) The power going into the satellite is
                                                        2 )T 4
                                                   σ(4πR⊙   ⊙1
                                             P =                 πD2 .
                                                      4πR2     4
         At equilibrium, this should equal to the power that the satellite emits.

                                                   P = σπD2 T 4 .

         Equating the two yields                            r
                                                                R⊙
                                                   T = T⊙
                                                                2R
     (b) The power that the satellite absorbs will be reduced by a factor of e0 , and the power emitted
         will also be reduced by a factor of e0 . When equating the powers as in part (a), the e0 will
         cancel out, giving the same answer.

     (c) As you can check from Planck’s law, the Sun has a higher intensity than the satellite at every
         frequency, but especially so for high frequencies, where the emission from the satellite falls
         off. Thus, it is better to have e(f ) small at these high frequencies, because energy comes in at
         these frequencies but doesn’t substantially come out. Thus, you could do something like this:

[3] Problem 19 (IPhO 1996). Two perfectly black surfaces of temperatures Th and Tℓ are parallel to
    each other in vacuum, and the net heat flux from the hotter surface to the colder one is P .

    Now suppose that N parallel, thermally insulating, perfectly black plates are placed in between
    them. This shielding reduces the heat flux to P ′ . Find P ′ in terms of P .

    Solution. Page 12 of the official solutions here does the problem for the case N = 2, getting the
    answer 1/3. An identical method works for general N , getting the answer 1/(N + 1).

  Example 10

  In problem 19 you considered some ideally black surfaces, but things get more complicated
  when the surfaces have general emissivity. For simplicity, consider two walls facing each other,
  with emissivities e1 and e2 . Verify that when both walls have temperature T , there is no net
  energy flow from one to the other.

  Solution

  Let I0 = σT 4 . The second wall emits radiation of intensity e2 I0 , so in equilibrium it must
  absorb this amount as well. Verifying this takes a little work.

  The first wall emits radiation of intensity e1 I0 , and a fraction e2 of it gets absorbed by the
  second wall. The rest bounces back to the first wall, and if it doesn’t get absorbed by the first
  wall, then it comes back to the second wall again for another chance to get absorbed. The
  total intensity transferred from the first wall to the second is the sum of an infinite series,
                                                                             e1 e2 I0
                 I1→2 = (e1 I0 ) (e2 + (1 − e2 )(1 − e1 )e2 + . . .) =                   .
                                                                         e1 + e2 − e1 e2
  This isn’t the expected result, because there’s another contribution. When the second wall
  emits radiation, it can reflect off the first wall and then get reabsorbed by the second wall.
  So the second wall absorbs a total radiation intensity due to its own emission, of

                                                                                (1 − e1 )e22 I0
         I2→2 = (e2 I0 ) ((1 − e1 )e2 + (1 − e1 )(1 − e2 )(1 − e1 )e2 + . . .) =                    .
                                                                                e1 + e2 − e1 e2
  When you sum these contributions, you get e2 I0 as expected. This is a typical example of
  the power of thermodynamics: it is often easy to tell what the answer has to be, but the way
  that answer comes about can be a bit complicated.

  Example 11

  Suppose a black sphere of radius r is placed inside a large container of arbitrary shape, whose
  walls are a blackbody with temperature T . In the steady state, what is the rate of power
  absorbed by the sphere from the walls?

  Solution
  Radiation is continually emitted from all points on the walls in all directions, so it would
  seem that we have to do a complicated integral to find how much hits the sphere, which
  would be hard even if we were given the container’s shape. But the problem is actually
  trivial due to the second law of thermodynamics!

  Let’s suppose the sphere itself is also at temperature T . Then it sends power (4πr2 )σT 4
  to the walls, since all radiation that exits the sphere hits the walls. But if the sphere and
  walls are at the same temperature, a temperature difference cannot spontaneously appear.
  Therefore, the power going from the walls to the sphere must also be (4πr2 )σT 4 . And

       since this quantity can also be expressed as a geometric integral, it can’t depend on the
       temperature of the sphere, so it must be true for any sphere temperature.

       For a tough problem which uses a variant of this idea, see EFPhO 2010, problem 2. An
       improved solution due to Stefan Ivanov is available here.


[User]
USAPhO 2004, problem B2. (The official solution is off by a sign in the last
    part. This classic setup also appears on USAPhO 2020, problem A1, and INPhO 2020, problem 2.
    However, the official solution to USAPhO 2020, problem A1 has factor of 2 errors.)

    4      Electromagnetic Waves
         Idea 8
         Maxwell’s equations have propagating wave solutions of the form
                                   E = E0 ei(k·r−ωt) ,    B = B0 ei(k·r−ωt)
         where E and B are in phase, perpendicular in direction, and have magnitudes E0 = cB0 .
         The propagation direction k is along E × B, and the wave speed is
                                                  ω      1
                                              c= = √          .
                                                  k     µ0 ϵ0

      Example 4

      Verify explicitly that in the absence of charges and currents, the electromagnetic field above
      satisfies Maxwell’s equations.

      Solution
      First let’s consider Gauss’s law, ∇ · E = 0. Splitting everything explicitly into components,

                               −iωt   ∂         ik·r    ∂        ik·r   ∂         ik·r
                      ∇·E=e              (E0,x e ) +      (E0,y e ) +      (E0,z e )
                                      ∂x               ∂y               ∂z

                               −iωt        ∂ ik·r         ∂ ik·r         ∂ ik·r
                            =e       E0,x e         + E0,y e      + E0,z e
                                          ∂x              ∂y            ∂z

                            = ei(k·r−ωt) (iE0,x kx + iE0,y ky + iE0,z kz )

                            = ik · E = 0

      since k is perpendicular to E0 . This is another example of a lesson we saw in M4. Namely,
      when everything is a complex exponential, differentiation is very easy. For an complex
      exponential in time, eiωt , differentiation with respect to time is just multiplication by iω.
      Similarly, for a field which is a complex exponential in space, eik·r , the divergence (∇·)
      becomes (ik·).

      By similar reasoning, Gauss’s law for magnetism is satisfied. Next, we check Ampere’s law,
                                                                  ∂E
                                              ∇ × B = µ0 ϵ0          .
                                                                  ∂t
      By the same logic as above, the curl becomes (ik×), while the time derivative becomes
      multiplication by −iω, giving
                                      ik × B = (−iω)µ0 ϵ0 E.
      Because k, E, and B are all mutually perpendicular, the directions of both sides match. Then
      all that remains is to check the magnitudes,

                                               kB0 = ωµ0 ϵ0 E0 .

      By plugging in results from above, this reduces to

                                                   c2 =
                                                          µ0 ϵ0

      which matches what we said above. (Or, if we didn’t know what c case, this logic would
      have been a way to derive it, as Maxwell did.) The verification of Faraday’s law is similar.
      Note that the displacement current term was essential; it wouldn’t have been possible to get
      electromagnetic wave solutions without it.

Solution:
USAPhO 2004, problem B2. (The official solution is off by a sign in the last
    part. This classic setup also appears on USAPhO 2020, problem A1, and INPhO 2020, problem 2.
    However, the official solution to USAPhO 2020, problem A1 has factor of 2 errors.)

    4      Electromagnetic Waves
         Idea 8
         Maxwell’s equations have propagating wave solutions of the form

                                     E = E0 ei(k·r−ωt) ,     B = B0 ei(k·r−ωt)

         where E and B are in phase, perpendicular in direction, and have magnitudes E0 = cB0 .
         The propagation direction k is along E × B, and the wave speed is
                                                      ω    1
                                                 c=     =√       .
                                                      k    µ0 ϵ0

      Example 4

      Verify explicitly that in the absence of charges and currents, the electromagnetic field above
      satisfies Maxwell’s equations.

      Solution
      First let’s consider Gauss’s law, ∇ · E = 0. Splitting everything explicitly into components,

                               −iωt   ∂         ik·r    ∂        ik·r   ∂         ik·r
                      ∇·E=e              (E0,x e ) +      (E0,y e ) +      (E0,z e )
                                      ∂x               ∂y               ∂z

                               −iωt        ∂ ik·r         ∂ ik·r         ∂ ik·r
                            =e       E0,x e         + E0,y e      + E0,z e
                                          ∂x              ∂y            ∂z

                            = ei(k·r−ωt) (iE0,x kx + iE0,y ky + iE0,z kz )

                            = ik · E = 0

      since k is perpendicular to E0 . This is another example of a lesson we saw in M4. Namely,
      when everything is a complex exponential, differentiation is very easy. For an complex
      exponential in time, eiωt , differentiation with respect to time is just multiplication by iω.
      Similarly, for a field which is a complex exponential in space, eik·r , the divergence (∇·)
      becomes (ik·).

      By similar reasoning, Gauss’s law for magnetism is satisfied. Next, we check Ampere’s law,
                                                                  ∂E
                                              ∇ × B = µ0 ϵ0          .
                                                                  ∂t
      By the same logic as above, the curl becomes (ik×), while the time derivative becomes
      multiplication by −iω, giving
                                      ik × B = (−iω)µ0 ϵ0 E.
      Because k, E, and B are all mutually perpendicular, the directions of both sides match. Then
      all that remains is to check the magnitudes,

                                               kB0 = ωµ0 ϵ0 E0 .

      By plugging in results from above, this reduces to

                                                   c2 =
                                                          µ0 ϵ0

      which matches what we said above. (Or, if we didn’t know what c case, this logic would
      have been a way to derive it, as Maxwell did.) The verification of Faraday’s law is similar.
      Note that the displacement current term was essential; it wouldn’t have been possible to get
      electromagnetic wave solutions without it.


[User]
Question B1
An AC power line cable transmits electrical power using a sinusoidal waveform with frequency
60 Hz. The load receives an RMS voltage of 500 kV and requires 1000 MW of average power.
For this problem, consider only the cable carrying current in one of the two directions, and ignore
effects due to capacitance or inductance between the cable and with the ground.

  a. Suppose that the load on the power line cable is a residential area that behaves like a pure
     resistor.

        i. What is the RMS current carried in the cable?
       ii. The cable has diameter 3 cm, is 500 km long, and is made of aluminum with resistivity
           2.8 × 10−8 Ω · m. How much power is lost in the wire?

  b. A local rancher thinks he might be able to extract electrical power from the cable using
     electromagnetic induction. The rancher constructs a rectangular loop of length a and width
     b < a, consisting of N turns of wire. One edge of the loop is to be placed on the ground; the
     wire is straight and runs parallel to the ground at a height h > a. Write the current in the
     wire as I = I0 sin ωt, and assume the return wire is far away.

        i. Determine an expression for the magnitude of the magnetic field at a distance r from
           the power line cable in terms of I, r, and fundamental constants.
       ii. Where should the loop be placed, and how should it be oriented, to maximize the induced
           emf in the loop?
       iii. Assuming the loop is placed in this way, determine an expression for the emf induced
            in the loop (as a function of time) in terms of any or all of I0 , h, a, b, N , ω, t, and
            fundamental constants.
       iv. Suppose that a = 5 m, b = 2 m, and h = 100 m. How many turns of wire N does the
           rancher need to generate an RMS emf of 120 V?

  c. The load at the end of the power line cable changes to include a manufacturing plant with
     a large number of electric motors. While the average power consumed remains the same, it
     now behaves like a resistor in parallel with a 0.25 H inductor.

        i. Does the power lost in the power line cable increase, decrease, or stay the same? (You
           need not calculate the new value explicitly, but you should show some work to defend
           your answer.)
       ii. The power company wishes to make the load behave as it originally did by installing a
           capacitor in parallel with the load. What should be its capacitance?

                   Copyright ©2011 American Association of Physics Teachers

2011 Semifinal Exam                            Part B                                              7

Solution:
Question B1
An AC power line cable transmits electrical power using a sinusoidal waveform with frequency
60 Hz. The load receives an RMS voltage of 500 kV and requires 1000 MW of average power.
For this problem, consider only the cable carrying current in one of the two directions, and ignore
effects due to capacitance or inductance between the cable and with the ground.

  a. Suppose that the load on the power line cable is a residential area that behaves like a pure
     resistor.

        i. What is the RMS current carried in the cable?
       ii. The cable has diameter 3 cm, is 500 km long, and is made of aluminum with resistivity
           2.8 × 10−8 Ω · m. How much power is lost in the wire?

  b. A local rancher thinks he might be able to extract electrical power from the cable using
     electromagnetic induction. The rancher constructs a rectangular loop of length a and width
     b < a, consisting of N turns of wire. One edge of the loop is to be placed on the ground; the
     wire is straight and runs parallel to the ground at a height h > a. Write the current in the
     wire as I = I0 sin ωt, and assume the return wire is far away.

        i. Determine an expression for the magnitude of the magnetic field at a distance r from
           the power line cable in terms of I, r, and fundamental constants.
       ii. Where should the loop be placed, and how should it be oriented, to maximize the induced
           emf in the loop?
       iii. Assuming the loop is placed in this way, determine an expression for the emf induced
            in the loop (as a function of time) in terms of any or all of I0 , h, a, b, N , ω, t, and
            fundamental constants.
       iv. Suppose that a = 5 m, b = 2 m, and h = 100 m. How many turns of wire N does the
           rancher need to generate an RMS emf of 120 V?

  c. The load at the end of the power line cable changes to include a manufacturing plant with
     a large number of electric motors. While the average power consumed remains the same, it
     now behaves like a resistor in parallel with a 0.25 H inductor.

        i. Does the power lost in the power line cable increase, decrease, or stay the same? (You
           need not calculate the new value explicitly, but you should show some work to defend
           your answer.)
       ii. The power company wishes to make the load behave as it originally did by installing a
           capacitor in parallel with the load. What should be its capacitance?

                                            Solution

                   Copyright ©2011 American Association of Physics Teachers

2011 Semifinal Exam                           Part B                                                 11

  a.    i. Because the load is purely resistive, the average power is simply

                                                 P = Irms Vrms

          so
                                                      P
                                            Irms =        = 2000 A.
                                                     Vrms
       ii. The cross-sectional area of the wire is A = πr2 = 7.07 × 10−4 m2 , so its resistance is
                                                     ρL
                                              R=        = 19.8 Ω.
                                                     A
          The power loss is then
                                            P = I 2 R = 79.2 MW.

  b.    i. The field is perpendicular to the wire and to the radius, and from Ampere’s Law
                                               I
                                                  B · ds = µ0 I.

          The integral evaluates to (2πr)B, giving
                                                          µ0 I
                                                     B=        .
                                                          2πr
          This well-known result can also be written down without justification.
       ii. The induced emf is proportional to the rate of change of the flux through the loop. Since
           the time dependence of the magnetic field is uniform across space, the rate of change of
           flux is maximized by maximizing the flux itself. This in turn can be accomplished by
           maximizing the field in the loop and ensuring that it is normal to the loop. Because
           the field gets stronger closer to the wire, the loop should be directly below the wire,
           and since the field is horizontal and perpendicular to the wire at this location, the loop
           should be vertical and parallel to the wire. Finally, again because the field gets stronger
           closer to the wire, the long edge of the loop should be vertical.
           In summary, the loop should be placed vertically, parallel to the wire and directly beneath
           it, with the long edge (of length a) vertical.
       iii. Faraday’s law states
                                                    dΦB
                                                  E =N
                                                      dt
          where we have dropped the sign, which is not important, and ΦB is the magnetic flux
          through a single loop. The flux is
                                             Z         Z h
                                     ΦB = B · dA = b       B(r) dr
                                                                   h−a

          where we have divided the loop into strips of radial width dr and length b. Plugging in
          the result of part (i),
                                         Z h
                                               µ0 I       µ0 Ib       h
                                  ΦB = b            dr =        log       .
                                           h−a 2πr         2π       h − a

                   Copyright ©2011 American Association of Physics Teachers

2011 Semifinal Exam                            Part B                                            12

          The time dependence comes only from the current,
                                                dI
                                                   = ωI0 cos ωt.
                                                dt
          Therefore, we have
                                            N µ0 b       h
                                        E=         log       I0 ω cos ωt.
                                             2π        h−a
       iv. The RMS value of I0 cos ωt is simply Irms , so taking the RMS value of both sides,
                                     N µ0 b      h                       h
                            Erms =          log     ωIrms = N µ0 bf log     Irms
                                      2π        h−a                     h−a
          where we used f = ω/2π. Plugging in the numbers, we have
                                                     h
                                        µ0 bf log       Irms = 0.0155 V
                                                    h−a
          so the required number of turns is
                                                     N = 7760.

  c.    i. The inductor adds a new component of the current in the wire which is 90◦ out of phase
           with the voltage. The rms current is increased, so the power lost in the wire increases.
           This can also be shown more formally using complex impedances. Ohm’s law is
                                                      V = IZ
          where both V and I are complex numbers, with their relative phase indicating the
          relative phase of the voltage and current. Without the inductance, Z = R. With the
          inductance, the two impedances add in parallel,
                                                       1 −1

                                                  1
                                           Z=       +         .
                                                  R iωL
           Since |Z| is lower and V remains the same, |I| increases, as argued above.
       ii. The most straightforward method is to use complex impedances. We wish to cancel the
           imaginary component supplied by the inductor, so we need
                                                 1
                                                     + iωC = 0
                                                iωL
          since they are in parallel. This is equivalent to
                                                           1
                                                     ω=√
                                                           LC
          so that
                                                    1
                                             C=        = 28.1 µF.
                                                  ω2L
          One can also arrive at this result without complex impedances. No external current is
          needed to make an LC circuit oscillate at its natural frequency; current simply sloshes
          back and forth between the inductor and capacitor, without any going through the
          power cable. Hence if the power cable frequency matches the natural frequency of the
          LC circuit, the LC circuit will not affect the current in the cable; the load
                                                                                √ behaves as if
          there were no LC circuit present at all. This gives the condition ω = 1/ LC as above.

                    Copyright ©2011 American Association of Physics Teachers

2011 Semifinal Exam                             Part B                                            13

USAPhO 2011 Question B1


[User]
3. Tyrannosaur (T. Rex) (6 points)

Paleontologists have
discovered tracks of a tyrannosaur where the footprints of the
same leg are A = 4.0 meters apart. They have also recovered
a piece of a tyrannosaur leg bone that has bone cross-sectionalo
areaN = 10000 times that ofa chicken (which the tyrannosaur
is relatedto).
i) (3 pt) Knowing that the chicken leg is approximately l =
15 cm tall, estimate the length of a tyrannosaur leg L. You may
assume that the length of a leg scales as the length of the whole
animal, and that the bone stress (force per area) is the same for
both animals. Isyourresultconsistent with the steplength A?
ii)(3pt)Estimatethenaturalwalkingspeedofthetyrannosaurby
approximatingthewalkingmotionofalegwithafreelyoscillating
pendulum motion. State clearlyall the assumptionsyou make.

Solution:
3. Tyrannosaur (T. Rex)

i. (3 pts) Knowing that mass m is proportional to volume, the
relationship between mass and length scale is L = l(M/m)1/3
.
The force F on animal bones is proportional to its mass and
to the cross-sectional area of the bone; hence, the area is pro-
potional to the mass. So, M
m = N, from which L = lN1/3
≈
3.23m. The step length is half of the distance between two
traces of the same leg, i.e. 2 m. This corrsponds to the angle
between the legs α = 2arcsin 1
3.2 ≈ 36◦
, which seems reason-
able.
ii. (3 pts) Let us model the leg with a physical pendulum.
The leg can be approximated as a uniform rod attached from
its upper end (the hip joint). Then its moment of inertia is
I = 1
3ML2
, where M is the leg mass.
For small-angle swings of the pendulum, the only force act-
ing on the leg is from the mass of the leg. Thus, the torque
equation will be
1
3
ML2
·φ̈ = −Mgφ
L
2
;
here, φ is the angle of the leg, so that the gravity force’s lever
arm is equal to φL
2 . Hence, the circular frequency of the leg
is ω =
√
3g
2L. The displacement A corresponds to the whole
period, i.e. the walking speed v = A/T = Aω/2π = A
2π
√
3g
2L ≈
1.2m/s ≈ 5km/h. So, the walking speed is comparable to that
of a human.
Problem

Tyrannosaur (T. Rex)


[User]
Given N independent measurements of the same quantity with the same uncertainty,     √
    xi ± ∆x, find the uncertainty of their sum. Hence show the uncertainty of their average is ∆x/ N .
        This result is extremely important, since repeating trials is one of the main ways to reduce
    uncertainty. But it’s important to remember that the results derived above hold only for independent
    measurements. For example, taking a single measurement, then averaging that single number with
    itself 100 times certainly wouldn’t reduce the uncertainty at all!

       Idea 3
       If x has uncertainty ∆x, and f (x) can be approximated by its tangent line, f (x′ ) ≈ f (x) +
       (x′ − x)f ′ (x) within the region x ± ∆x, then f (x) has approximate uncertainty f ′ (x) ∆x.
                                                                                √

Solution:
The uncertainty of their sum ∆X can be found by adding in quadrature,
                                           v
                                           uN
                                           uX           √
                                    ∆X = t (∆xi )2 = N ∆x.
                                                   i=1

    Therefore, the uncertainty of the average is
                                                 ∆X  ∆x
                                                    =√ .
                                                 N    N

       Idea 3
       If x has uncertainty ∆x, and f (x) can be approximated by its tangent line, f (x′ ) ≈ f (x) +
       (x′ − x)f ′ (x) within the region x ± ∆x, then f (x) has approximate uncertainty f ′ (x) ∆x.

                                                                                √


[User]
2. Pendulum (7 points)

Consider an elastic rod, the mass and the compressibility of which (i.e.
the length change) can be neglected in this problem. It can be assumed
that if one end of the rod is firmly fixed, and a force F is applied to the
other end of the rod, perpendicularly to the rod at the point of applica-
tion, then the rod takes a form of a circle segment. The radius of that
circle is inversely proportional to the force, R = k/F, where the factor
k is a characteristic of the rod.
R
F
1)Letthe rod befixedvertically, atits bottomend,andaball ofmassm
beattachedtoitsupperend. Knowing thefactork,thelengthoftherod
l, and the free fall acceleration g, find the period of small oscillations of
the ball. In this question, youmay assumethat gml fi k.
2)Whatis themaximal massM oftheball, which canbestably heldon
such a vertical rod?
Remark: you may use approximate expressions sinx ≈ x − x3
/6
and cosx ≈ 1 − x2
/2 (for x fi 1).

Solution:
2. Pendulum

1) The smallness of the oscillations means that the angle of the arc for-
med by the rod is small, α = l/R fi 1. Therefore, the force returning
the ball is almost horizontal, F = k/R = kα/l, and the horizontal
displacement of the ball is x = R(1 − cosα) ≈ Rα2
/2 = lα/2, hence
α = 2x/l.
For horizontalequation of motion,we can neglect the gravity force,
which is of the order of mgα fi αk/l; the right-hand side here hap-
pens to be the elastic force. So, mẍ = −kα/l = −2kx/l2
, hence
ω =
p
2k/m/l; the period T = 2π/ω = πl
p
2m/k.
2) Elastic energy of the rod depends apparently only on its shape, i.e.
on the curvature radius R, or, equivalently, on the binding angle α.
First, we derive the expression for that energy: Πr =
Rx
0
Fdx =
R
(2kx/l2
)dx = kx2
/l2
= kα2
/4. This is to be compared with the
changeofthegravitationalpotentialenergyΠg = −mg(l−Rsinα) =
−mgl(1−sinα/α) ≈ −mglα2
/6. So,the vertical position is stable, if
kα2
/4 − mglα2
/6 > 0, i.e. 3k > 2mgl.
Remark: a simple-minded force balance of the ball to the direction,
perpendicular to the rod leads to another result k > mgl. This is not
correct, because if we bind the rod with a force applied to its tip, small
displacements of the tip are not perpendicular to the rod (as one might
think). In particular, this means that if there are both perpendicular
force F and tangential tensionT in the rod, the curvature radius R will
depend also on the tension T. In our solution, we avoided such kind
of complications by noting that the elastic energy depends only on the
shape of the rod (if we bind the rod with some force F while keeping
T = 0, we don’tneed to bother about this effect).

Pendulum


[User] Jahipüss massiga $M = \SI{3,5}{kg}$ ripub horisontaalselt kahe paralleelse niidi otsas. Tulistamisel kerkis jahipüss $h = \SI{20}{cm}$ võrra tagasilöögi tõttu. Kuuli mass $m = \SI{10}{g}$. Määrata kuuli kiirus $v_0$ torust väljalennu hetkel.

Solution:
Püssi kiiruse $u_{0}$ vahetult peale tulistamist saame leida impulsi jäävuse seadusest:
$M u_{0}=m v_{0}$, kus $v_{0}$ on otsitav kuuli algkiirus. Peale tulistamist alustab püss niidi otsas pendlisarnasat liikumist. Oma algse kineetilise energia $M u_{0}^{2} / 2$ kulutab püss tõusmiseks kõrgusele $h$, kus tema potentsiaalne energia on $M g h$. Energia jäävuse tõttu:
$$
\frac{1}{2} M u_{0}^{2}=M g h \quad \Rightarrow \quad u_{0}=\sqrt{2 g h}.
$$
Siit
$$
v_{0}=\frac{M u_{0}}{m}=\frac{M}{m} \sqrt{2 g h} \approx \SI{693}{m/s}.
$$

Jahipüss

Jahipüss massiga $M = \SI{3,5}{kg}$ ripub horisontaalselt kahe paralleelse niidi otsas. Tulistamisel kerkis jahipüss $h = \SI{20}{cm}$ võrra tagasilöögi tõttu. Kuuli mass $m = \SI{10}{g}$. Määrata kuuli kiirus $v_0$ torust väljalennu hetkel.


[User]
Some questions about forces between dipoles and other multipoles.

     (a) Above, you’ve shown that the force between permanent magnetic dipoles falls off as 1/r4 .
         How about two permanent electric dipoles?

     (b) How about a permanent dipole and a permanent quadrupole?

        (c) How about two permanent quadrupoles?

     (d) Now consider an ion and a neutral atom. The electric field of the ion polarizes the atom; the
         field of that induced dipole then reacts on the ion. Show that the resulting force is attractive
         and falls as 1/r5 .

    4     Electromagnetic Waves in Matter
    In this section, you will work out some of the theory of electromagnetic waves in matter.

         Idea 9
         In the absence of any free charge or current, Maxwell’s equations in matter are identical to
         Maxwell’s equations in vacuum, except that ϵ0 and µ0 are related by ϵ and µ, so the waves
                                  √
         propagate with speed 1/ ϵµ = c/n, with E = (c/n)B.

Solution:
(a) The basic form of the fields and forces is identical, so the answer is the same.

     (b) The field of a quadrupole goes like 1/r4 , so energy of the dipole goes like U ∼ mB ∼ 1/r4 .
         Thus, the interaction energy in this case goes like 1/r4 , for a force of 1/r5 .

        (c) A single quadrupole is two dipoles with moments m and −m separated by dr where the
            magnitude of the quadrupole moment is ∼ |m||dr|. (Technically, the quadrupole moment is a
            tensor, and the sizes of its individual components depends on the relative orientation of m
            and dr, but we won’t worry about that detail here, since we’re only looking for the scaling of
            the force with r.) The energy of the quadrupole is

                         m · B(r + dr) − m · B = m · ((some sort of derivative of B) · dr).

           The field of one quadrupole is 1/r4 , so its derivative is 1/r5 . Thus the energy of interaction
           goes like 1/r5 , for a force of 1/r6 .

     (d) The field of the ion falls as 1/r2 , so the dipole moment induced is p ∼ 1/r2 . Furthermore, the
         dipole moment points along the field and hence the displacement between the ion and atom,
         indicating the force is attractive. The electric field from the dipole (and hence the force) goes
         as p/r3 ∼ 1/r5 . (You might wonder if the induced dipole then gives the ion itself a dipole
         moment. It does, but the resulting force is much weaker than the one we found here, between
         the induced dipole and the ion’s overall charge.)

    4      Electromagnetic Waves in Matter
    In this section, you will work out some of the theory of electromagnetic waves in matter.

       Idea 9
       In the absence of any free charge or current, Maxwell’s equations in matter are identical to
       Maxwell’s equations in vacuum, except that ϵ0 and µ0 are related by ϵ and µ, so the waves
                                √
       propagate with speed 1/ ϵµ = c/n, with E = (c/n)B.


[User]
Richard Feynman used to tell the following story, here reproduced verbatim.
            I was in the cafeteria and some guy, fooling around, throws a plate in the air. As the
            plate went up in the air I saw it wobble, and I noticed the red medallion of Cornell on
            the plate going around. It was pretty obvious to me that the medallion went around
            faster than the wobbling.
           I had nothing to do, so I start figuring out the motion of the rotating plate. I discover
           that when the angle is very slight, the medallion rotates twice as fast as the wobble rate
           – two to one. It came out of a complicated equation!
            I went on to work out equations for wobbles. Then I thought about how the electron
            orbits start to move in relativity. Then there’s the Dirac equation in electrodynamics.
            And then quantum electrodynamics. And before I knew it... the whole business that I
            got the Nobel prize for came from that piddling around with the wobbling plate.
    Feynman was right about quantum electrodynamics, but was he right about the plate?

    5      Gravity
[3] Problem 21 (Morin 5.65). Let the Earth’s radius be R, its average density be ρ, and its angular
    frequency of rotation be ω. Consider a long rope with uniform mass density extending radially from
    just above the surface of the Earth out to a radius ηR. Show that if the rope is to remain above
    the same point on the equator at all times, then we must have
                                                            8πGρ
                                                 η2 + η =         .
                                                             3ω 2

    What is the numerical value of η, and where does the tension in the rope achieve its maximum value?
    Such a rope would function as a space elevator, allowing objects to be lifted to space much more
    cheaply. It was conceived by the science fiction writer R. A. Heinlein, who called it a “skyhook”.

[2] Problem 22 (Morin 10.7). A puck slides with a small speed v on frictionless ice. The surface
    is “level” in the sense that it is orthogonal to geff at all points, where geff includes the centrifugal
    acceleration. Show that the puck moves in a circle, as seen in the Earth’s rotating frame. Find
    its radius and the angular frequency and direction of the motion, in terms of the Earth’s angular
    velocity ω0 and the latitude ϕ of the puck.

Solution:
For concreteness, take the angular momentum of the plate to point upward. From the
problem statement, the axis of symmetry of the plate is a small angle θ away from this direction.

As in M8, we decompose the angular momentum into parallel and perpendicular components, and

                                          L∥ = L,     L⊥ = θL

by the small angle approximation, and hence
                                    L∥     L                     L⊥     θL
                             ω∥ =      =         ,        ω⊥ =      =
                                    I∥   M R2 /2                 I⊥   M R2 /4

where the last step is by the perpendicular axis theorem. Now we need to think more about the
physical motion of the plate. The component ω∥ of angular velocity parallel to the axis of rotation
is the part that makes the medallion go around,

                                              ωmed = ω∥ .

The component ω⊥ makes the orientation of the plate itself rotate. Specifically, the entire setup
drawn above rotates about the axis of L with angular velocity ωwob . Imagine the path taken by
the unit normal n̂ to the plate. The tip of this vector goes in a circle of circumference 2πθ, but the
speed of the tip of the vector is ω⊥ . Therefore, the angular velocity of the vector along the circle is
                                                          ω⊥
                                             ωwob =          .
                                                           θ
The answer to the question is
                                              ωmed  1
                                                   = .
                                              ωwob  2
So it’s the opposite of what Feynman says – actually the wobbling goes twice as fast. Sometimes,
when you tell a story too many times, you forget details like this.

    5    Gravity
[3] Problem 21 (Morin 5.65). Let the Earth’s radius be R, its average density be ρ, and its angular
    frequency of rotation be ω. Consider a long rope with uniform mass density extending radially from
    just above the surface of the Earth out to a radius ηR. Show that if the rope is to remain above
    the same point on the equator at all times, then we must have
                                                           8πGρ
                                              η2 + η =           .
                                                            3ω 2
    What is the numerical value of η, and where does the tension in the rope achieve its maximum value?
    Such a rope would function as a space elevator, allowing objects to be lifted to space much more
    cheaply. It was conceived by the science fiction writer R. A. Heinlein, who called it a “skyhook”.

    Solution. The gravitational field from Earth will be

                                              g = GπρR3 /r2

    which works with the tension to provide the centripetal acceleration ω 2 r. For a small piece of rope
    of length dr and mass dm = µ dr, the force balance gives

                                           ω 2 r dm = g dm − dT

                                              4 GπρµR3
                                       dT =            dr − µω 2 r dr
                                              3   r2
    Integrating from r = R to r = ηR gives

                                         4     2    1    1
                         T (ηR) − T (R) = GπρµR 1 −     − µω 2 R2 (η 2 − 1).
                                         3          η    2

    At both ends of the rope, the tension must be zero since they’re not connected to anything, so
                                        8πGρ η − 1
                                                   = (η − 1)(η + 1)
                                         3ω 2  η
    which gives
                                                      8πGρ
                                           η2 + η =         = 579
                                                       3ω 2
    and solving the quadratic numerically gives

                                                  η = 23.6.

    The maximum value of the tension occurs when dT /dr = 0, which is when
                                                                     1/3
                                   4πGρR3

                                3                            4πGρ
                               r =        ,       r=R                       = 6.62R.
                                     3ω 2                     3ω 2

    This radius has a physical meaning: since the gravitational and centrifugal forces on a piece of mass
    balance here, it’s the radius where a satellite can stay in geostationary orbit.

[2] Problem 22 (Morin 10.7). A puck slides with a small speed v on frictionless ice. The surface
    is “level” in the sense that it is orthogonal to geff at all points, where geff includes the centrifugal
    acceleration. Show that the puck moves in a circle, as seen in the Earth’s rotating frame. Find
    its radius and the angular frequency and direction of the motion, in terms of the Earth’s angular
    velocity ω0 and the latitude ϕ of the puck.

    Solution. Since the surface is level with gravity and the centrifugal acceleration, the normal force
    will cancel out the effects from those, so the only remaining force is the Coriolis force −2mΩ × v.
    The component of Earth’s angular velocity normal to the ground at latitude ϕ is Ω sin ϕ, so
                                                            v
                           2ω0 v sin ϕ = v 2 /r,    r=             ,   ω = 2ω0 sin ϕ.
                                                         2ω0 sin ϕ
    The puck will travel clockwise in the Northern hemisphere and counterclockwise in the Southern
    hemisphere. (You might wonder why this is opposite the direction hurricanes turn. The difference
    is that in a hurricane, the center has low pressure, and the Coriolis force provides a outward force
    which opposes the inward pressure force, so that the system doesn’t immediately collapse. By
    contrast, here the Coriolis force must be inward since it is the only source of centripetal force.)


[User]
To check that you understand example 6, let the cylinder axis be vertical, and
    suppose the solenoid wires are wound at an angle θ to the horizontal. The example corresponds
    to θ = 0◦ , but when θ = 90◦ the wires run straight up, leading to an inward pressure at the walls.
    Find the angle θ at which the pressure vanishes.

Solution:
To avoid juggling too many variables, we’ll just work in terms of the surface current
    density K. At the walls, we have

                                        K = K0 sin θ ẑ + K0 cos θ θ̂

    where K0 is some constant that depends on the wire spacing. (Note that K0 is not equal to nI,
    since the wires are at an angle to the horizontal. But it won’t affect the answer since everything
    will scale with K0 .) Applying Ampere’s law gives
                                                  (
                                                    cos θ ẑ inside
                                     B = µ0 K0 ×                     .
                                                    sin θ θ̂ outside

    By the same logic as in the example, we have an outward pressure (µ0 K02 /2) cos2 θ, but now we
    also have an inward pressure (µ0 K02 /2) sin2 θ. The two cancel when θ = 45◦ .


[User]
Physics Competition

Theoretical Competition

Please read this first:
The time available is 5 hours for the theoretical competition.
1. Use only the pen provided.
2. Use only the one side of the paper.
3. Begin each part of the problem on a separate sheet.
4. For each question, in addition to the blank sheets where you may write, there is an
    answer sheet where you must summarize the results you have obtained.
    Numerical results should be written with as many digits as are appropriate to the
    given data.
5. Write on the blank sheets of paper whatever you consider is required for the
    solution of the question. Please use as little text as possible; express your answers
    primarily in equations, numbers, figures, and plots.
6. Fill in the boxes at the top of each answer sheet of paper used by writing your
    student code as shown on your identification tag, and additionally on the “blank”
    sheets: your student code, the problem number, the progressive number of each
    sheet (Page n. from 1 to N) and the total number (N) of “blank” sheets that you
    use and wish to be evaluated (Page total) for each problem; If you use some
    blank sheets of paper for notes that you do not wish to be marked, put a large X
    across the entire sheet and do not include it in your numbering.
7. At the end of the competition, arrange all sheets for each problem in the following
    order:

    (a) answer sheet
    (b) used sheets in order
    (c) the sheets you do not wish to be marked
    (d) unused sheets and the printed questions

    Place the papers inside the envelope provided and leave everything on your desk.
    You are not allowed to take any sheets of paper out of the room.

Theoretical problem 1
Back-and-Forth Rolling of a Liquid-Filled Sphere (10 points)
      Consider a sphere filled with liquid inside rolling back and forth at the bottom
of a spherical bowl. That is, the sphere is periodically changing its translational and
rotational direction. Due to the viscosity of the liquid inside, the movement of the
sphere would be very complicated and hard to deal with. However, a simplified model
presented here would be beneficial to the solution of such a problem.
     Assume that a rigid thin spherical shell of radius r and mass m is fully filled with
some liquid substance of mass M, denoted as W. W has such a unique property that
usually it behaves like an ideal liquid (i.e. without any viscosity), while in response to
some special external influence (such as electric field) it transits to solid state
immediately with the same volume; and once the applied influence removed, the
liquid state recovers immediately. Besides, this influence does not give rise to any
force or torque exerting on the sphere. This liquid-filled spherical shell (for
convenience, called ‘the sphere’ hereafter) is supposed to roll back and forth at the
bottom of a spherical bowl of radius R ( R > r ) without any relative slipping, as shown
in the figure. Assume the sphere moves only in the vertical plane (namely, the plane
of the figure), please study the movement of the sphere for the following three cases:
1. W behaves as in ideal solid state, meanwhile W contacts the inner wall of the
   spherical shell so closely that they can be taken as solid sphere as a whole of
   radius r with an abrupt density change across the interface between the inside wall
   of the shell and W.
   (1) Calculate the rotational inertia I of the sphere with respect to the axis passing
   through its center C. (You are asked to show detailed steps.)
                                                                            (1.0 points)
   (2) Calculate the period T1 of the sphere rolling back and forth with a small
   amplitude without slipping at the bottom of the spherical bowl.           (2.5points)
                                                    D

                                                             R
                                                    θ ’0 θ 0
                                                                     C
                                A‘0                                  A
                                      A‘1                           A 0
                                                                  A2 1
                                            solid     O    liquid

                                      liquid                     solid

2. W behaves as an ideal liquid with no friction between W and the spherical shell.
   Calculate the period T2 of the sphere rolling back and forth with a small
   amplitude without slipping at the bottom of the spherical bowl.         (2.5 points)
3. W transits between ideal solid state and ideal liquid state.
   Assume at time t = 0 , the sphere is kept at rest, the line CD makes an angle θ 0

   ( θ 0 << 1 rad）with the plumb line OD, where D is the center of the spherical bowl.

   The sphere contacts the inner wall of the bowl at point A0 , as shown in the figure.
   Release the sphere, it starts to roll left from rest. During the motion of the sphere
   from A0 to its equilibrium position O, W behaves as ideal liquid. At the moment
   that the sphere passes through point O, W changes suddenly into solid state and
   sticks itself firmly on the inside wall of the sphere shell until the sphere reaches
   the left highest position A0' . Once the sphere reaches A0' , W changes suddenly
   back into the liquid state. Then, the sphere rolls right; and W changes suddenly
   into solid state and sticks itself firmly on the inside wall of the spherical shell
   again when the sphere passes through the equilibrium position O. When the
   sphere reaches the right highest position A1 , W changes into liquid state once
   again. Then the whole circle repeats time after time. The sphere rolls right and left
   periodically but with the angular amplitude decreased time after time. The motion
   direction of the sphere is shown by curved arrows in the figure, together with the
   words “solid” and “liquid” showing corresponding state of W. It is assumed that
   during such process of rolling back and forth, no any relative slide happens
   between the sphere and the inside wall of the bowl (or, alternatively, the bottom of
   the bowl can supply as enough friction as needed). Calculate the period T3 of the

   sphere rolling right and left, and the angular amplitude θ n of the center of the
   sphere, namely, the angle that the line CD makes with the vertical line OD when
   the sphere reaches the right highest position An for the n-th time (only A2 is

   shown in the figure). （4.0 points）

Solution:
Solution of the theoretical problem 1

Back-and-Forth Rolling of a Liquid-Filled Sphere

1.1 Let         and      denote the rotational inertia of the spherical shell and W in

solid state respectively, while I be the sum of and        . The surface mass density of

the spherical shell is              . Cut a narrow zone from the spherical shell

perpendicular to its diameter, which spans a small angle             with respect to the
center of the sphere C, while the spherical zone makes an angle         with the diameter
of the spherical shell, which is called C axes hereafter, as shown in Fig. 1. The

rotational inertia of the narrow zone about the C axis is                               ,

therefore integral over the whole spherical shell gives

                                                                             (1A.1)

                                                                            dα r
                                                                               α
                                                                            C

                                                                          Figure 1

The volume density of W is                    . By using above result for the spherical

zone it can be seen that the rotational inertia of the solid W about the C axis is

                                                                             (1A.2)

Then,                                                                          (1A.3)

2According to the Newton’s second law we can derive the translational motion
equation of the center of mass for the sphere along the tangent of the bowl,

                                                                               (1A.4)

                                                                  D

                                                                      θ
                                                                              R

                                                                          E
                                                                                C     f
                                                                 C            ϕ
                                                                  O
                                                                          (m+M)g
                                                                  Figure 2

where             denotes the angular position of the center of mass of the sphere as

shown in Fig.2, and f is the frictional force acting on the sphere by the inside wall of
the bowl. From the rotational dynamics, we have,

                                                                                (1A.5)

where ϕ is the angular position of the reference radius CE with respect to the starting

position. Assumed constraint of pure rolling on the motion of the sphere reads,

                                                                                (1A.6)

Equations (1A.4)-(1A.6) lead to

This is a motion equation of the type of simple harmonic oscillator. Therefore, we
obtain the angular frequency and period of the sphere rolling right and left:

                                                                                (1A.7)

                                         .                                            (1A.8)

2. This case can be treated similarly, except taking that the ideal liquid does not rotate
into consideration. Therefore Eqs. (1A.4) and (1A.6) are still applicable, while Eq.
(1A.5) needs to be modified as

                                                                                (1A.9)

Equations (1A.4), (1A.6), and (1A.9) result in

Then, the angular frequency and period of the sphere rolling back-and-forth are
obtained respectively.

                                                                                (1A.10)

                                                                                (1A.11)

3. The time taken by the sphere from position A0 to equilibrium position O is             ,

      from O to       , and      from       to O,         from O to     . Although the

angular amplitude decreases step by step (see below) during the rolling process of the
sphere right and left, the period keeps unchanged. This means

                                                                                (1A.12)

     Next, we calculate the change of the angular amplitude. When the sphere passes

 through the equilibrium position O after it rolled down from the initial position        ,

 the velocity of its center is

                                                                                (1A.13)

Now the angular velocity of the spherical shell rotating about the C axis is

                                                                                (1A.14)

where C axis is the axis of rotation through the center of the sphere and perpendicular
to the paper plane of Fig.2. When W behaves as liquid (before it changes into solid
state), the angular momentum of the sphere relative to point O is

                                                                               (1A.15)

When W changes suddenly into solid state, due to the fact that both gravitational and
 frictional force pass through point O, the angular momentum of the sphere relative to
 O is conserved, we have

                                                                               (1A.16)

where      and      represent the angular velocity of the sphere immediately before
and after passing through point O. Therefore

                                                                          (1A.17)

while after passing through point O the velocity of the center of the sphere becomes

                                                                          (1A.18)

Once the sphere reaches the left highest position           corresponding to the left

angular amplitude      we have

However,

From above two expressions we obtain

                                                                          (1A.19)

Similarly we can treat the process that the sphere rolls from position    back to A1 ,

the second highest position on the right, corresponding to the second right angular

amplitude θ1, and obtain

Then,

Following the similar procedure repeatedly we finally obtain:

                                                                          (1A.20)

APhO 2007 Theory T1


[User]
An infinite solenoid with radius b has n turns per unit length. The current varies
    in time according to I(t) = I0 cos ωt. A ring with radius r < b and resistance R is centered on the
    solenoid’s axis, with its plane perpendicular to the axis.

      (a) What is the induced current in the ring?

     (b) A given little piece of the ring will feel a magnetic force. For what values of t is this force
         maximum? At this moment, sketch the electric field everywhere.

        (c) What is the effect of the force on the ring? That is, does the force cause the ring to translate,
            spin, etc.?

     (d) If the current is driven by the AC power from a wall outlet, which has frequency 60 Hz in
         America, the ring will emit a humming sound. What is the frequency of this sound?

Solution:
An infinite solenoid with radius b has n turns per unit length. The current varies
    in time according to I(t) = I0 cos ωt. A ring with radius r < b and resistance R is centered on the
    solenoid’s axis, with its plane perpendicular to the axis.

     (a) What is the induced current in the ring?

     (b) A given little piece of the ring will feel a magnetic force. For what values of t is this force
         maximum? At this moment, sketch the electric field everywhere.

     (c) What is the effect of the force on the ring? That is, does the force cause the ring to translate,
         spin, etc.?

     (d) If the current is driven by the AC power from a wall outlet, which has frequency 60 Hz in
         America, the ring will emit a humming sound. What is the frequency of this sound?

    Answer. (c) grows and shrinks, (d) 120 Hz


[User]
6. CUBE (5 points)

— Taavet Kalda. A laser
pointer of power P is directed at a glass cube.
Therefractiveindexofthecubeis n. Thesurface
of the cube has anti-reflective coating so that
there is no partial reflection when light travels
from one medium to the other. The speed of
light is c.
i) (3 points) What is the maximum force by
which the laser pointer can push the cube if the
laser needs to be kept parallel to one of the faces
of the cube (this means that the laser beam can
propagate only in a two-dimensional plane)?
ii) (2 points) What is the maximum force by
which the laser pointer can push the cube if the
orientation of the laser can be arbitrary?

Solution:
6. CUBE The cube gets pushed by the light

reflecting against its surfaces. Since there is no
partialreflection,lightcanonlyreflectinsidethe
cube via total internal reflection.
Let the cube’s faces be aligned to x-y-z axis
and let the light enter from the face which is
perpendicular to the z-axis.
Before entering the cube, let the unit vec-
tor directed along the motion of the light
be ⃗ t = (tx,ty,tz), after entering the cube,
⃗ r = (rx,ry,rz), before leaving the cube, ⃗ r′
=
(r′
x,r′
y,r′
z) and after leaving the cube, ⃗ t′
=
(t′
x,t′
y,t′
z). Every time the light bounces against
one of the sides of the cube, the respective com-
ponent of⃗ r gets flipped.
i) The laser beam is limited to propagate in a
two-dimensional plane. Take ty = 0, ry = 0,
r′
y =0 and t′
y =0.
In time dt, the laser pointer generates light
with total energy Pdt carrying momentum P
c dt.
In that time, the same amount of light enters
the cube and exits it, only with different dir-
ection. Applying Newton’s III law, the cube
mustattainamomentumofd⃗ p = P
c dt(⃗ t−⃗ t′
)and
thus experiences a force of ⃗ F =
d⃗ p
dt = P
c (⃗ t−⃗ t′
) =
P
c
√
(tx −t′
x)2 +(tz −t′
z)2. This means that we
wish to maximize the quantity (tx −t′
x)2
+(tz −
t′
z)2
.
Snell’s law can be written as nrx = tx and
nr′
x = t′
x since tx and rx are the sines of angles
of incidence and departure respectively.
The laser beam can only reflect against the
side that is perpendicular to the x-axis, beam
path with internal reflection is shown inthe fig-
ure. Thus, r′
z = rz and t′
z = tz. Let’s investigate
the reflectionagainst the x-face. The angle of in-
cidence is cosα= rx. The condition for total in-
ternal reflection is sinαn ≥ 1. Rearranging the
terms yields cosα <
√
1− 1
n2 or rx <
√
1− 1
n2 .
This means that tx <
p
n2 −1.
θr
θr θr
α α
The force is maximal when the laser beam
bounces against the cube odd number of times.
Then r′
x = −rx and t′
x − tx < 2
p
n2 −1. Thus,
F = P
c
√
(tx −t′
x)2 +(tz −t′
z)2 < 2P
c (n2
− 1).
Note that t2
x +t2
z =1 so tx <1. This means that
the force can’t be larger than 2P
c . The maximal
force is then
F =
{
2P
c
p
n2 −1, if n<
p
2
2P
c , otherwise
ii) We proceed in a similar way as in the pre-
vious part, the main difference being that the
y-component doesn’t have to be 0.
The act of entering the cube keeps the light
moving in the same direction in the x-y plane.
Thus, tx
ty
=
rx
ry
. Snell’s law can be written
as
√
t2
x +t2
y = n
√
r2
x +r2
y, since
√
t2
x +t2
y and
√
r2
x +r2
y are the sines of the angles of incidence
and departure respectively. Combining these
equations, we get rx =
tx
n , ry =
ty
n . Similarly,
t′
x = nr′
x and t′
y = nr′
y.
The act of reflecting against the sides of the
cube doesn’t change the magnitude of rx and
ry. Thus, t′
z = tz. This means that the quant-
ity (tx −t′
x)2
+(ty −t′
y)2
needs to be maximized
and this happens when r′
y = −ry and r′
x = −rx
so F =2P
c
√
t2
x +t2
y =2nP
c
√
r2
x +r2
y.
Using the same argumentation as in the pre-
vious subtask, the condition for a reflection to
happen against the x-face is tx <
p
n2 −1. Sim-
ilarly, ty <
p
n2 −1 must hold for the y-face.
This means that r2
x+r2
y <2
(
1− 1
n2
)
. On the
otherhand, t2
x+t2
y+t2
z =1 so t2
x+t2
y <1 and r2
x+
r2
y < 1
n2 . Thus, r2
x +r2
y < min
(
2
(
1− 1
n2
)
, 1
n2
)
=
1
n2 min
(
2
(
n2
−1
)
,1
)
. The maximum force the
cube can experience is then
F =
{
2
p
2P
c
p
n2 −1, if n<
p
3/2
2P
c , otherwise

CUBE


[User]
Consider a mass m on a table attached to a spring at the origin with zero relaxed
    length, which exerts the force
                                           F = −kr
    on the mass. We will find the general solution for r(t) = (x(t), y(t)) in two different ways.

      (a) Directly write down the answer, using the fact that the x and y coordinates are independent.

     (b) Sketch a representative sample of solutions. What kind of curve does the trajectory follow?

      (c) ⋆ Here’s a more unusual way to arrive at the same answer. Go to a noninertial reference frame
          rotating with angular velocity ω0 about the origin, so that the centrifugal force cancels out
          the spring force. In this frame, the only relevant force is the Coriolis force −2mω0 × v. Find
          the general solution in this frame, then transform back to the original frame and show that
          you get the same answer as in part (a). (This can get a bit messy; the easiest way is to treat
          the plane as the complex plane, i.e. work in terms of the variable r = x + iy.)

    4    Optimal Launching
    Finally, we’ll consider projectile motion questions that involve optimization. These are rare on the
    USAPhO, but they are quite fun problems, with occasionally very slick solutions.

        Example 8

        A bug wishes to jump over a cylindrical log of radius R lying on the ground, so that it just
        grazes the top of the log horizontally as it passes by. What is the minimum launch speed v
        required to do this?

        Solution
        Let P be the point at the top of the log. For the bug to be moving horizontally at P , energy
        conservation applied to the vertical motion gives an initial vy obeying

                                         1                           p
                                           mv 2 = 2mgR,        vy = 2 gR.
                                         2 y
        Thus, we need to find the minimum vx for the motion to be possible. If vx is too low, the
        hypothetical trajectory of the bug will instead pass through the log. At the lowest possible
        vx , the bug’s trajectory is not just tangent to the log at point P , but also has the same radius
        of curvature (i.e. the trajectory and the log’s shape have the same first and second derivatives).

        For uniform motion in a circle of radius r, the acceleration is a = v 2 /r. Conversely, when an
        object follows a trajectory of instantaneous radius of curvature r, its acceleration component
        normal to the path must be a = v 2 /r. So applying this to the bug at P gives

                                                  vx2            p
                                             g=       ,   vx =    gR.
                                                  R
        Thus, the minimum initial speed is
                                                 q           p
                                            v=    vx2 + vy2 = 5gR.

        This radius of curvature trick doesn’t come up often, but it’s cool when it does.

Solution:
(a) We just have two separate equations for each component,
                                         d2 x    k         d2 y    k

                                              = − x,          2
                                                                = − y.
                                         dt      m         dt      m
                                                                                p
     Both describe a harmonic oscillator with angular frequency ω0 =             k/m. Then the general
     solution can be written as

                             x(t) = A cos(ω0 t + ϕ1 ),     y(t) = B sin(ω0 t + ϕ2 ).

     In general, it is very rare for the x and y coordinates to be independent. Another example of
     this type is projectile motion in linear drag, F = −kv. In these cases the 2D or 3D problem
     is no harder than the 1D version, but we’re rarely so lucky.

 (b) In the case where ϕ1 = ϕ2 = 0 and A = B, the mass moves in a circle centered at the origin.
     More generally, when the angles ϕi are unequal, the mass can move in an ellipse with center
     at the origin.
                                                                                p
 (c) The centrifugal force is mω02 r, so to cancel the spring force we need to choose ω0 = k/m.
     Now, in the rotating frame, the Coriolis force acts just like a magnetic field: it’s always
     perpendicular to the motion, so the solution is circular motion. The angular frequency ωc of
     that circular motion satisfies
                                                    mv 2
                                          2mω0 v =       = mωc v
                                                     r
     from which we conclude ωc = 2ω0 . So in complex notation,

                                              r(t) = r0 + r1 e2iω0 t

     in the rotating frame. We can return to the original frame by simply multiplying by e−iω0 t ,
     to give
                                      r(t) = r0 e−iω0 t + r1 eiω0 t .
     Taking real and imaginary parts and letting ri = ai + ibi ,

     x(t) = (a0 + a1 ) cos(ω0 t) + (b0 − b1 ) sin(ω0 t),   y(t) = (b0 + b1 ) cos(ω0 t) + (a1 − a0 ) sin(ω0 t).

     This is the same as our result for part (a), after you use the sine and cosine addition formulas
     and appropriately redefine the parameters. Evidently, elliptical motion is just the superposition
     of two opposite circular motions! (In general, complex numbers are a useful way to deal with
     magnetic or Coriolis forces for motion in a plane, where B or ω points perpendicular to the
     plane. In these cases the force lies in the plane perpendicular to the velocity, so it’s just
     proportional to iṙ, which is nice and simple; we’ll see this idea again later.)

    4    Optimal Launching
    Finally, we’ll consider projectile motion questions that involve optimization. These are rare on the
    USAPhO, but they are quite fun problems, with occasionally very slick solutions.

        Example 8

        A bug wishes to jump over a cylindrical log of radius R lying on the ground, so that it just
        grazes the top of the log horizontally as it passes by. What is the minimum launch speed v
        required to do this?

        Solution
        Let P be the point at the top of the log. For the bug to be moving horizontally at P , energy
        conservation applied to the vertical motion gives an initial vy obeying

                                         1                           p
                                           mv 2 = 2mgR,        vy = 2 gR.
                                         2 y
        Thus, we need to find the minimum vx for the motion to be possible. If vx is too low, the
        hypothetical trajectory of the bug will instead pass through the log. At the lowest possible
        vx , the bug’s trajectory is not just tangent to the log at point P , but also has the same radius
        of curvature (i.e. the trajectory and the log’s shape have the same first and second derivatives).

        For uniform motion in a circle of radius r, the acceleration is a = v 2 /r. Conversely, when an
        object follows a trajectory of instantaneous radius of curvature r, its acceleration component
        normal to the path must be a = v 2 /r. So applying this to the bug at P gives

                                                  vx2            p
                                             g=       ,   vx =    gR.
                                                  R
        Thus, the minimum initial speed is
                                                 q           p
                                            v=    vx2 + vy2 = 5gR.

        This radius of curvature trick doesn’t come up often, but it’s cool when it does.