[User]
01^      USAPhO 2021, problem B1. An elegant rotation problem.

[3] Problem 15 (Morin 8.24). A spherically symmetric ball of radius R initially slides without rotating
    on a horizontal surface with friction. The initial speed is v0 , and the moment of inertia about the
    center is I = βmR2 .

      (a) Assuming that the normal force is always applied upward at the bottom of the ball, and
          that the friction force is always applied horizontally at the bottom of the ball (but assuming
          nothing about how the friction force varies), find the speed of the ball when it begins to roll
          without slipping. Also, find the kinetic energy lost while sliding.

     (b) Now consider the case where the friction force is standard uniform kinetic friction, f = −µk N .
         Verify that the work done by friction equals the energy loss calculated in part (a).

        (c) In reality, the conclusions above can be modified by “rolling resistance”. Any real material
            will slightly deform when the ball rolls on it. We can crudely account for this by thinking
            of the normal force as applied not at the bottom of the ball, but at a point slightly forward
            from the bottom. The horizontal component of this normal force is defined to be fr = −µr Ny
            where Ny is the vertical normal force, and µr ≪ 1. In addition, kinetic friction is still present,
            as in part (b). Under these assumptions, find the velocity of the ball once it stops slipping.
            Is more or less energy lost than in part (b)?

       Remark
       In the early 1800s, some said it was impossible for a train engine to pull anything heavier
       than the engine itself. As the argument went, the force that moves the train forward is
       friction between the engine car’s wheels and the track. If the engine car has mass M , the
       maximum friction force is µM . If the rest of the train has mass M ′ , however, then it requires
       a force µM ′ to get it started moving, so the train can’t start if M ′ > M .

       Problem 15 explains why this reasoning is wrong. The maximum forward frictional force on
       the engine car wheels is determined by the coefficient of static friction µs , while the force
       needed to get the rest of the train moving is determined by the coefficient of rolling friction
       µr . So we only need µs M > µr M ′ , which is possible since µr can be very small. For steel
       train wheels on steel rail, we might have µs ∼ 0.5 but µr ≲ 10−3 .

       On the other hand, early trains could have trouble going up inclines. This led to several
       innovative concepts, such as trains propelled by atmospheric pressure or pushed by mechanical
       legs. All the mechanical systems we’re familiar with today, whose design might seem obvious
       at first glance, actually gradually evolved through many intermediate forms. For instance,
       most people think they know how a bicycle works, but actually don’t, because it’s really
       quite tricky. Accordingly, it took over a century for the modern bicycle design to evolve.

[4] Problem 16 (IPhO 1998). A hexagonal pencil with mass M and side length R is pushed and
    rolls down a ramp of inclination θ. For some values of θ, the pencil will roll down the plane with
    some terminal speed, never losing contact with the ramp. In order to avoid a complicated moment
    of inertia calculation, we will assume the cross section looks like a wheel with six equally spaced
    massless spokes and no rim, with all the mass on the axle.

      (a) The pencil does not speed up indefinitely, but rather reaches a steady state. Explain why, and
          compute the speed the pencil’s axis has immediately after each collision, in the steady state.

     (b) Find the minimum θ so that rolling spontaneously starts, without needing a push.

      (c) Find the minimum θ so that, once the pencil has been pushed to start rolling, it never stops.

     (d) Find the maximum θ so that a rolling pencil always remains in contact with the plane.

Solution:
(a) The point here is that we can relate the linear and angular impulses without having
     to know how the force behaves in time. If there’s a net
                                                          R impulse J on the ball, the net change
     in angular momentum about the center of the ball is RF dt = RJ. When the ball is rolling
     without slipping, v = ωR. Thus

                                     J = mv0 − mvf = L/R = βmRω

     which gives
                                                          v0
                                                 vf =        .
                                                         1+β
     The kinetic energy lost is
                   1      1      1          1                       1 β
               ∆K = mv02 − mvf2 − βmR2 ω 2 = m(v02 − (1 + β)vf2 ) =      mv02 .
                   2      2      2          2                       21+β

 (b) Here, f = −µk mg and acts for a time of t = J/f . Since the acceleration is constant,
     the ball travels a distance of 12 (v0 + vf )t while sliding. However, it will turn a distance of
     Rθ = 12 ωRt = 12 vf t in the other direction, so the relative distance traveled between the surface
     of the ball and the ground is L = 12 v0 t.

                                     1      1                1 β
                           ∆K = f L = v0 J = m(v0 − vf )v0 =      mv02
                                     2      2                21+β
     as desired.

 (c) Now the angular and linear accelerations are
                                            µk g
                                       α=        ,   a = −(µr + µk )g
                                            βR
     where the rolling resistance doesn’t affect the angular acceleration, because the overall normal
     force always exerts no torque about the center of mass of the ball. (Note that this conclusion
     would have been changed if we accounted for the deformation of the ball itself, which would
     give a second, additional type of rolling resistance. Here we are assuming that the ball is much
     harder than the surface it rolls on, though there are plenty of situations where the reverse is
     true, such as when a bike tire rolls on concrete.)
     Thus, by similar reasoning to that of part (a),
                                                        v0
                                          vf =                      .
                                                 1 + β(1 + µr /µk )

     This is smaller than the result of part (a), so more energy is lost. The reason is that the
     rolling resistance dissipates additional energy. Notice that even once slipping stops, rolling
     resistance will continue to dissipate energy, causing the ball to eventually come to a stop.

   Remark
   In the early 1800s, some said it was impossible for a train engine to pull anything heavier
   than the engine itself. As the argument went, the force that moves the train forward is
   friction between the engine car’s wheels and the track. If the engine car has mass M , the
   maximum friction force is µM . If the rest of the train has mass M ′ , however, then it requires

       a force µM ′ to get it started moving, so the train can’t start if M ′ > M .

       Problem 15 explains why this reasoning is wrong. The maximum forward frictional force on
       the engine car wheels is determined by the coefficient of static friction µs , while the force
       needed to get the rest of the train moving is determined by the coefficient of rolling friction
       µr . So we only need µs M > µr M ′ , which is possible since µr can be very small. For steel
       train wheels on steel rail, we might have µs ∼ 0.5 but µr ≲ 10−3 .

       On the other hand, early trains could have trouble going up inclines. This led to several
       innovative concepts, such as trains propelled by atmospheric pressure or pushed by mechanical
       legs. All the mechanical systems we’re familiar with today, whose design might seem obvious
       at first glance, actually gradually evolved through many intermediate forms. For instance,
       most people think they know how a bicycle works, but actually don’t, because it’s really
       quite tricky. Accordingly, it took over a century for the modern bicycle design to evolve.

[4] Problem 16 (IPhO 1998). A hexagonal pencil with mass M and side length R is pushed and
    rolls down a ramp of inclination θ. For some values of θ, the pencil will roll down the plane with
    some terminal speed, never losing contact with the ramp. In order to avoid a complicated moment
    of inertia calculation, we will assume the cross section looks like a wheel with six equally spaced
    massless spokes and no rim, with all the mass on the axle.
      (a) The pencil does not speed up indefinitely, but rather reaches a steady state. Explain why, and
          compute the speed the pencil’s axis has immediately after each collision, in the steady state.

     (b) Find the minimum θ so that rolling spontaneously starts, without needing a push.

      (c) Find the minimum θ so that, once the pencil has been pushed to start rolling, it never stops.

     (d) Find the maximum θ so that a rolling pencil always remains in contact with the plane.
    Solution. (a) Each time the pencil rolls through an angle π/3, a new vertex of the pencil hits
         the plane. In this moment, that vertex suddenly becomes the new pivot point about which
         the pencil is rotated, which means energy is dissipated in an inelastic collision. This is the
         reason that the pencil reaches a steady state, instead of accelerating indefinitely. You can see
         this very nicely depicted in this video.
          Let the pencil’s center of mass have speed vi just before an impact, and vf just after the
          impact. The impact involves a singular impact force at the new pivot point, which means
          angular momentum is conserved about that point. Thus,

                                             M vi R cos 60◦ = M vf R

          from which we conclude
                                                           vi
                                                     vf =     .

          In the steady state, the kinetic energy gained from rolling from one vertex to another bal-
          ances the energy lost in the collision, so conserving energy immediately after a collision and
          immediately before a next one gives
                                         1                     1
                                           M vf2 + M gR sin θ = M (2vf )2
                                         2                     2

     from which we conclude                          r
                                                          2gR sin θ
                                              vf =                  .

     By the way, the original formulation of the question gave the pencil a moment of inertia
     CM R2 about its center of mass. The solution with general C is not much harder. Now the
     angular momentum conservation condition is

                               M vi R cos 60◦ + CM R2 ωi = (C + 1)M R2 ωf

     where ωi = vi /R and ωf = vf /R. Thus,

                                                      2C + 1 vi
                                               vf =             .
                                                       C +1 2
     Likewise, the energy balance equation for the steady state becomes
                             1                            1
                               M (C + 1)vf2 + M gR sin θ = M (C + 1)(2vf )2
                             2                            2
     from which we conclude
                                       s
                                                 2gR sin θ(C + 1/2)2
                                vf =                                        .
                                           (C + 1) ((C + 1)2 − (C + 1/2)2 )

 (b) This is a basic statics problem. The rolling must start if the center of mass of the hexagon is
     not above its support, which implies θ = 30◦ .

 (c) Between two vertex transitions, the maximum potential energy of the pencil is when the
     center of mass is directly above the vertex at a height R. It will start out at a height of
     h0 = R sin(θ + 60◦ ) above the vertex, and fall down to a height hf = R sin(60◦ − θ) above the
     vertex.
     In order for it to roll indefinitely, potential energy from height R to hf followed by the inelastic
     collision must leave enough kinetic energy for the pencil to go from height h0 to height R.
     Earlier we found that ωf = ω0 (C + 1/2)/(C + 1), so the kinetic energy will be reduced by a
     factor of α = (ωf /ω0 )2 . Thus the energy equation for indefinite rolling is

                                       αM g(R − hf ) = M g(R − h0 ).

     This implies
                                             1 − sin(θ + 60◦ )
                                                               = α.
                                             1 − sin(60◦ − θ)
     In our case, α = 1/4. The solution of the above equation can be found using either binary
     search or by the “plug in” method, i.e. repeatedly calculating

                                            1 − sin(60◦ − Ans)

                                arcsin 1 −                       − 60◦ .

     Both methods give an answer of θ = 10.21◦ .

     (d) The pencil leaves the ramp when gravity isn’t strong enough to provide the needed centripetal
         acceleration for the rotation about a vertex. Right before the next vertex transition, the pencil
         is moving the fastest, and the radial component of gravity is the smallest, so the pencil most
         readily leaves the ramp at that point. Using part (a)’s notation (ω0 is the angular velocity
         right before the next transition), the leaving condition is g cos(30◦ + θ) = g sin(60◦ − θ) = ω02 R,
         where 30◦ + θ is the angle between the vertical and line connecting the center of mass to the
                                                  v
         vertex. Using our expression for ω0 = Rf (C + 1)/(C + 1/2) found in part (a),
                                                            2 sin θ(C + 1)
                                      sin(60◦ − θ) =
                                                       ((C + 1)2 − (C + 1/2)2 )
         With C = 0, we have
                                                            8 sin θ
                                               sin(60◦ − θ) =       .

         We can binary search for the answer or repeatedly plug in
                                                  3 sin(60◦ − Ans)

                                         arcsin

          to find that the maximum angle for it to stay on the ramp is θ = 15.3◦ . So the range of angles
          where the rolling will never stop, but also keep the pencil on the ramp, is quite narrow!
    This famous question has appeared on the IPhO, BAUPC, and Morin’s mechanics book, and papers
    have even experimentally confirmed its results. If you would like to know even more about it, see

                   01m
    the extended analysis here.

                   01^


[User]
Theory
                                                    English (Official)                        Q1-1
Two Problems in Mechanics (10 points)
Please read the general instructions in the separate envelope before you start this problem.

Part A. The Hidden Disk (3.5 points)
We consider a solid wooden cylinder of radius r1 and thickness h1 . Somewhere inside the wooden cylin-
der, the wood has been replaced by a metal disk of radius r2 and thickness h2 . The metal disk is placed
in such a way that its symmetry axis B is parallel to the symmetry axis S of the wooden cylinder, and
is placed at the same distance from the top and bottom face of the wooden cylinder. We denote the
distance between S and B by d. The density of wood is ρ1 , the density of the metal is ρ2 > ρ1 . The total
mass of the wooden cylinder and the metal disk inside is M .
In this task, we place the wooden cylinder on the ground so that it can freely roll to the left and right.
See Fig. 1 for a side view and a view from the top of the setup.
The goal of this task is to determine the size and the position of the metal disk.
In what follows, when asked to express the result in terms of known quantities, you may always assume
that the following are known:

                                                    r1 , h 1 , ρ1 , ρ2 , M .                               (1)

The goal is to determine r2 , h2 and d, through indirect measurements.

                 a)                                           b)                    S   B
                                                                               r1       r2

                                S                                              h1            h2
                                       d       r2
                               r1
                                           B

                                    Figure 1: a) side view b) view from above

We denote b as the distance between the centre of mass C of the whole system and the symmetry axis
S of the wooden cylinder. In order to determine this distance, we design the following experiment: We
place the wooden cylinder on a horizontal base in such a way that it is in a stable equilibrium. Let us now
slowly incline the base by an angle Θ (see Fig. 2). As a result of the static friction, the wooden cylinder
can roll freely without sliding. It will roll down the incline a little bit, but then come to rest in a stable
equilibrium after rotating by an angle φ which we measure.

                                               Theory
                                            English (Official)                 Q1-2
                                                  S
                                                  φ

                                      Θ

                                 Figure 2: Cylinder on an inclined base.

    A.1     Find an expression for b as a function of the quantities (1), the angle φ and the   0.8pt
            tilting angle Θ of the base.

From now on, we can assume that the value of b is known.

                                                  S
                                                  φ

                                     Figure 3: Suspended system.

Next we want to measure the moment of inertia IS of the system with respect to the symmetry axis S.
To this end, we suspend the wooden cylinder at its symmetry axis from a rigid rod. We then turn it away
from its equilibrium position by a small angle φ, and let it go. See figure 3 for the setup. We find that φ
describes a periodic motion with period T .

                                                Theory
                                             English (Official)                  Q1-3
    A.2     Find the equation of motion for φ. Express the moment of inertia IS of the            0.5pt
            system around its symmetry axis S in terms of T , b and the known quantities
            (1). You may assume that we are only disturbing the equilibrium position by a
            small amount so that φ is always very small.

From the measurements in questions A.1 and A.2, we now want to determine the geometry and the
position of the metal disk inside the wooden cylinder.

    A.3     Find an expression for the distance d as a function of b and the quantities (1).      0.4pt
            You may also include r2 and h2 as variables in your expression, as they will be
            calculated in subtask A.5.

    A.4     Find an expression for the moment of inertia IS in terms of b and the known           0.7pt
            quantities (1). You may also include r2 and h2 as variables in your expression,
            as they will be calculated in subtask A.5.

    A.5     Using all the above results, write down an expression for h2 and r2 in terms of       1.1pt
            b, T and the known quantities (1). You may express h2 as a function of r2 .

Part B. Rotating Space Station (6.5 points)
Alice is an astronaut living on a space station. The space station is a gigantic wheel of radius R rotating
around its axis, thereby providing artificial gravity for the astronauts. The astronauts live on the inner
side of the rim of the wheel. The gravitational attraction of the space station and the curvature of the
floor can be ignored.

    B.1     At what angular frequency ωss does the space station rotate so that the astro-        0.5pt
            nauts experience the same gravity gE as on the Earth's surface?

Alice and her astronaut friend Bob have an argument. Bob does not believe that they are in fact living
in a space station and claims that they are on Earth. Alice wants to prove to Bob that they are living on
a rotating space station by using physics. To this end, she attaches a mass m to a spring with spring
constant k and lets it oscillate. The mass oscillates only in the vertical direction, and cannot move in the
horizontal direction.

    B.2     Assuming that on Earth gravity is constant with acceleration gE , what would be       0.2pt
            the angular oscillation frequency ωE that a person on Earth would measure?

    B.3     What angular oscillation frequency ω does Alice measure on the space station?         0.6pt

Alice is convinced that her experiment proves that they are on a rotating space station. Bob remains
sceptical. He claims that when taking into account the change in gravity above the surface of the Earth,
one finds a similar effect. In the following tasks we investigate whether Bob is right.

                                                Theory
                                            English (Official)                  Q1-4
                                                                        ωss
                                                           R

                                         Figure 4: Space station

    B.4     Derive an expression of the gravity gE (h) for small heights h above the surface     0.8pt
            of the Earth and compute the oscillation frequency ω̃E of the oscillating mass
            (linear approximation is enough). Denote the radius of the Earth by RE . Neglect
            the rotation of Earth.

Indeed, for this space station, Alice does find that the spring pendulum oscillates with the frequency that
Bob predicted.

    B.5     For what radius R of the space station does the oscillation frequency ω match        0.3pt
            the oscillation frequency ω̃E on the Earth? Express your answer in terms of RE .

Exasperated with Bob's stubbornness, Alice comes up with an experiment to prove her point. To this end
she climbs on a tower of height H over the floor of the space station and drops a mass. This experiment
can be understood in the rotating reference frame as well as in an inertial reference frame.
In a uniformly rotating reference frame, the astronauts perceive a fictitious force FC⃗ called the Coriolis
force. The force FC⃗ acting on an object of mass m moving at velocity v ⃗ in a rotating frame with constant
angular frequency ω⃗ss is given by

                                            FC⃗ = 2mv ⃗ × ω⃗ss .                                         (2)

In terms of the scalar quantities you may use

                                           FC = 2mvωss sin φ ,                                           (3)

where φ is the angle between the velocity and the axis of rotation. The force is perpendicular to both the
velocity v and the axis of rotation. The sign of the force can be determined from the right-hand rule, but
in what follows you may choose it freely.

                                                Theory
                                             English (Official)                  Q1-5
    B.6     Calculate the horizontal velocity vx and the horizontal displacement dx (relative     1.1pt
            to the base of the tower, in the direction perpendicular to the tower) of the
            mass at the moment it hits the floor. You may assume that the height H of
            the tower is small, so that the acceleration as measured by the astronauts is
            constant during the fall. Also, you may assume that dx ≪ H.

To get a good result, Alice decides to conduct this experiment from a much taller tower than before. To
her surprise, the mass hits the floor at the base of the tower, so that dx = 0.

    B.7     Find a lower bound for the height of the tower for which it can happen that           1.3pt
            dx = 0.

Alice is willing to make one last attempt at convincing Bob. She wants to use her spring oscillator to show
the effect of the Coriolis force. To this end she changes the original setup: She attaches her spring to
a ring which can slide freely on a horizontal rod in the x direction without any friction. The spring itself
oscillates in the y direction. The rod is parallel to the floor and perpendicular to the axis of rotation of
the space station. The xy plane is thus perpendicular to the axis of rotation, with the y direction pointing
straight towards the center of rotation of the station.

                                                                y=0

                                                            d

                                             Figure 5: Setup.

    B.8     Alice pulls the mass a distance d downwards from the equilibrium point x = 0,         1.7pt
            y = 0, and then lets it go (see figure 5).
                • Give an algebraic expression of x(t) and y(t). You may assume that ωss d is
                  small, and neglect the Coriolis force for motion along the y-axis.
                • Sketch the trajectory (x(t), y(t)), marking all important features such as
                  amplitude.

                                      Theory
                                   English (Official)   Q1-6
Alice and Bob continue to argue.

Solution:
Problem 1 : Solution/marking scheme  Two Problems in Mechanics (10
points)

Part A. The Hidden Disk (3.5 points)

 A1 (0.8 pt) Find an expression for b as a function of the quantities (1), the angle φ and
 the tilting angle Θ of the base.

 Solution A1:                                                                                [0.8]

 Geometric solution:     use that torque with respect to point of contact is 0       ⇒ cen-
 ter of gravity has to be vertically above point of contact.

                                                     D
                                           sin φ =                                              0.3
                                                     b
                                                     D
                                           sin Θ =                                              0.3
                                                     r1
 Here D may be called another name. Solve this:

                                      r1             r1 sin Θ
                            sin φ =      sin Θ ⇒ b =                                            0.2
                                       b              sin φ

 Alternative: Torque and forces with respect to another point:                                  [0.8]
 Correct equation for torque                                                                    0.3
 Correct equation for force                                                                     0.3
 Correct solution                                                                               0.2

 A2 (0.5 pt) Find the equation of motion for φ. Express the moment of inertia IS of the
 cylinder around its symmetry axis S in terms of T , b and the known quantities (1). You
 may assume that we are only disturbing the equilibrium position by a small amount so
 that φ is always very small.

 Solution A2:                                                                                [0.5]

                     Write some equation of the form φ̈ = −ω
                                                                 2
                                                                     φ                          0.1
 Writing an equation of the form φ = A cos ωt is also correct.
 Two solutions:
                         1      2
    1. Kinetic energy:
                         2 IS φ̇ and potential energy:    −bM g cos φ.   Total energy is con-
       served, and dierentiation w.r.t. time gives the equation of motion.

    2. Angular equation of motion from torque, τ = IS φ̈ = −M gb sin φ.

 Correct equation (either energy conservation or torque equation of motion)                     0.3
 Final answer                         s
                                           IS         M gbT 2
                           T = 2π              ⇒ IS =                                           0.1
                                          M gb         4π 2

                                                 1

(Derivation:
                                           bM g           bgM
                                ⇒ φ̈ = −        sin φ ≃ −     φ
                                            IS             IS
so that
                                                   bgM
                                            ω2 =
                                                    IS
)

A3 (0.4 pt) Find an expression for the distance d as a function of b and the quantities
(1).   You may also include r2 and h2 as variables in your expression, as they will be
calculated in subtask   A.5.

Solution A3:                                                                              [0.4]

Some version of the center of mass equation, e.g.

                                              dM2
                                       b=                                                 0.2
                                            M 1 + M2
correct solution:
                                                bM
                                   d=                                                     0.2
                                         πh2 r22 (ρ2 − ρ1 )

A4 (0.7 pt) Find an expression for the moment of inertia IS in terms of b and the known
quantities (1). You may also include r2 and h2 as variables in your expression, as they
will be calculated in subtask   A.5.

Solution A4:                                                                              [0.7]

correct answer for moment of inertia of homogeneous disk

                                           1
                                       I1 = πh1 ρ1 r14                                    0.2
                                           2
Mass wrong                                                                                -0.1
Factor 1/2 wrong in formula for moment of inertia of a disk                               -0.1
Correct answer for moment of inertia of `excess' disk:

                                      1
                                  I2 = πh2 (ρ2 − ρ1 )r24                                  0.2
                                      2
Using Steiner's theorem:

                           IS = I1 + I2 + d2 πr22 h2 (ρ2 − ρ1 )                           0.1
correct solution:

                    1            1                        b2 M 2
                IS = πh1 ρ1 r14 + πh2 (ρ2 − ρ1 )r24 + 2                                   0.2
                    2            2                   πr2 h2 (ρ2 − ρ1 )

                                                 2

 In terms of d rather than b gives 0.1pts rather than 0.2pts for the final answer:              0.1
                         1            1
                     IS = πh1 ρ1 r14 + πh2 (ρ2 − ρ1 )r24 + d2 πr22 h2 (ρ2 − ρ1 )
                         2            2

 A5 (1.1 pt) Using all the above results, write down an expression for h2 and r2 in terms
 of b, T and the quantities (1). You may express h2 as a function of r2 .

 Solution A5:                                                                                [1.1]

 It is not clear how exactly students will attempt to solve this system of equations. It is
 likely that they will use the following equation:

                                 M = πr12 h1 ρ1 + πr22 h2 (ρ2 − ρ1 ) .                         0.3
                 2
 solve IS for r2 :

                                                                       M2

                               2                   1
                 r22 =                         IS − πh1 ρ1 r14 − b2                            0.4
                          M − πr12 h1 ρ1           2                M − πr12 h1 ρ1

 replace IS by T :
                                                      M gbT 2
                                               IS =                                            0.1
                                                       4π 2
 solve correctly for r2 :

                     s
                                             bgT 2 1                   M2

                              2
            r2 =                           M                 4
                                                   − πh1 ρ1 r1 − b2                            0.1
                         M − πr12 h1 ρ1       4π 2  2               M − πr12 h1 ρ1

                                                                     2          2
 write down an equation for h2 along the lines of M = πr1 ρ1 h1 + πr2 (ρ2 − ρ1 )h2 and solve
 it correctly:
                                                  M − πr12 ρ1 h1
                                           h2 =                                                0.2
                                                  πr22 (ρ2 − ρ1 )

Part B. Rotating Space Station (6.5 points)

 B1 (0.5 pt) At what angular frequency ωss does the space station rotate so that the
 astronauts experience the same gravity gE as on the Earth's surface?

 Solution B1:                                                                                [0.5]

 An equation for the centrifugal force along the lines of

                                                Fce = mω 2 r                                   0.1

                                                        3

Balancing the forces, correct equation

                                             2
                                       gE = ωss R                                         0.2
Correct solution
                                                                                0.2
                                             p
                                     ωss =    gE /R

B2 (0.2 pt) Assuming that on Earth gravity is constant with acceleration gE , what
would be the angular oscillation frequency ωE that a person on Earth would measure?

Solution B2:                                                                              [0.2]

                         Realize that result is independent of gE                         0.1
Correct result:
                                                                                0.1
                                              p
                                      ωE =     k/m

B3 (0.6 pt) What angular oscillation frequency ω does Alice measure on the space
station?

Solution B3:                                                                              [0.6]

some version of the correct equation for force

                                               2
                                   F = −kx ± mωss x                                       0.2
getting the sign right
                                               2
                                   F = −kx + mωss x                                       0.2
Find correct dierential equation

                                              2
                                 mẍ + (k − mωss )x = 0                                   0.1
Derive correct result
                                                                                0.1
                                         p
                                    ω=           2
                                          k/m − ωss
                           2
Using gE /R instead of ωss is also correct.

B4 (0.8 pt) Derive an expression of the gravity gE (h) for small heights h above the
surface of the Earth and compute the oscillation frequency ω̃E (linear approximation is
enough). The radius of the Earth is given by RE .

                                               4

Solution B4:                                                                                [0.8]

                              gE (h) = −GM/(RE + h)2                                           0.1
linear approximation of gravity:

                                        GM     GM
                           gE (h) = −    2 + 2h 3 + . . .                                      0.2
                                        RE     RE

Realize that gE = GM/RE :
                          2

                           gE (h) = −gE + 2hgE /RE + . . .                                     0.1
Opposite sign is also correct, as long as it is opposite in both terms.
Realize what this means for force, i.e. that the constant term can be eliminated by shifting
the equilibrium point:
                               F = −kx + 2xmgE /RE                                             0.2
Find correct dierential equation

                              mẍ + (k − 2mgE /RE )x = 0                                       0.1
correct result
                                                                                0.1
                                       p
                               ω̃E =       k/m − 2gE /RE

No points are deducted if student answers with ω̃E /(2π) because oscillation frequency
might also be interpreted as inverse period.

B5 (0.3 pt) For what radius R of the space station does the oscillation frequency ω
match the oscillation frequency ω̃E on the surface of the Earth? Express your answer in
terms of RE .

Solution B5:                                                                                [0.3]

Write down equation
                                     2
                                    ωss = 2gE /RE                                              0.1
Solve
                                       R = RE /2                                               0.2
         2
If GM/RE rather than gE is used, give only         0.1pt.

B6 (1.1 pt) Calculate the horizontal velocity vx and the horizontal displacement dx
(relative to the base of the tower, in the direction perpendicular to the tower) of the mass
at the moment it hits the floor. You may assume that the height H of the tower is small,
so that the acceleration as measured by the astronauts is constant during the fall. Also,
you may assume that dx fi H .

                                               5

Solution B6:                                                                               [1.1]

There are several possible solutions.
Solution one  Using Coriolis force
   • Velocity vx
     Equation for Coriolis force with correct velocity:

                                          2
                              FC (t) = 2mωss            3
                                             Rtωss = 2mωss Rt                              0.1
     Integrate this, or realize that it is like uniform acceleration for the velocity:

                                                   3
                                         vx (t) = ωss Rt2                                  0.2
     plug in correct value for
                                                                                0.2
                                             p
                                        t=           2 R
                                                 2H/ωss
     overall correct result
                                          vx = 2Hωss                                       0.1
   • The displacement dx :
     Integrate vx (t):
                                             1  3 3
                                         dx = Rωss t                                       0.3
                                             3
                                                                           1
     Instead of integrating, students may simply `average' by taking
                                                                           2 of the final
     velocity. This gives a factor of
                                      1
                                      2 instead of . Deduct a total of 0.1 pts for this.
                                                  1
                                                  3                                        -0.1

     Plug in value for t
                                                                     r
                    1                    1                   1           8H 3
                       3
                dx = Rωss      2
                          (2H/ωss R)3/2 = 23/2 H 3/2 R−1/2 =                               0.2
                    3                    3                   3            R
Solution two  Using inertial frame This solution is similar to the way to solve B7,
but needs more complicated approximations than Solution one.

   • vx
     Here φ denotes the angle swept by the mass and α the angle the astronauts (and
     tower) has rotated when the mass lands on the floor, see

     Initially the velocity of the mass in an inertial frame is vx = ωss (R − H).          0.1
     When the mass lands, the x-direction has been rotated by φ so the new horizontal
     velocity component is then

                                        ωss (R − H) cos φ                                  0.1
     (Student may also write cos α instead of cos φ, since dx fi H .)

                                             R−H     H
                                 cos φ =         =1−                                       0.1
                                              R      R
     Transforming to the rotating reference frame, one needs to subtract ωss R.            0.1
     Finally in the reference frame of the astronauts

                          H 2

                                                    H
           vx = ωss R 1 −     − ωss R ≈ ωss R 1 − 2     − ωss R = −2ωss H                  0.2
                          R                         R
     The sign of the velocity depend on the choice of reference direction, so a positive
     sign is also correct.

                                                 6

   • dx
     With the notation from the calculation of vx

                                       dx = (α − φ)R                                               0.1

                                                      H
                                       φ = arccos 1 −
                                                      R
                                              α = ωss t
     where t is the fall time of the mass, which is given by

                                        p
                                            R2 − (R − H)2
                                  t=                                                               0.1
                                            ωss (R − H)

     (see solution to B7)

     Writing ξ ≡ H/R this means

                              "p                            #
                                1 − (1 − ξ)2
                         dx =                − arccos(1 − ξ) R                                     0.3
                                  1−ξ

     which is a valid end answer to the problem.           It is possible, but not necessary, to
     approximate this for small ξ :

                                               p        ξ
                                arccos(1 − ξ) ≈ 2ξ 1 +
                                                       12
     which after insertion into the equation for dx and approximation of small ξ yields
     the same result as in Solution one:
                                                     r
                                                 2       2H 3
                                            dx =
                                                 3        R

     If this end answer misses the factor 2/3, deduct 0.1 points.                                  -0.1

Solution three  Inertial frame with geometry trick
This is an alternative solution to obtain dx
The mass travels the distance l, and during the fall the space station rotates by φ, see
Figure 2. According to the intersecting chord theorem,

                                  l2 = H(2R − H)                                                   0.1
The rotated angle is φ = ωss t where

                                             l
                                       t=                                                          0.1
                                            R−H
is the fall time. Thus                 p
                                           H(2R − H)
                                 φ=                                                                0.1
                                            R−H
                                      p
              d             l          H(2R − H)
                                                                                0.1
                                                         p
                = φ − arcsin =                   − arcsin x(2 − x)
              R             R           R−H

                                               7

                           Figure 1: Notation for solution two

                          Figure 2: Notation for solution three.

                           p
Denote x ≡ H/R and y ≡         x(2 − x). Since

                                                      y3
                                     arcsin y ≈ y +
                                                      6
one gets

                                                                            r
          d                                                     √       2       2H 3
            ≈ y(1 + x) − y − y 3 /6 = y(x − y 2 /6) ≈ 2xy/3 ≈ 2x 2x/3 =
          R                                                             3        R

Final answer                                                                                0.1

B7 (1.3 pt) Find a lower bound for the height of the tower for which it can happen that
dx = 0.

Solution B7:                                                                                [1.3]

The key is to use a non-rotating frame of reference. If the mass is released close enough

                                             8

to the center, its linear velocity will be small enough for the space station to rotate more
than 2π before it hits the ground.
The velocity is given by
                                     v = ωss (R − H)                                           0.1
distance d that the mass flies before hitting the space station

                                 d2 = R2 − (R − H)2                                            0.1
use non-rotating frame of reference to obtain time t until impact
                                        p
                                         R2 − (R − H)2
                              t = d/v =                                                        0.1
                                          ωss (R − H)
Now there are several possible ways to relate H and the rotated angle φ of the space
station:
Solution one
                                            R sin φ
                                     t=                                                        0.2
                                           ωss R cos φ
This time must match t = φ/ωss . Obtain the equation

                                         φ = tan φ                                             0.2
Realizing that there is an infinite number of solutions.                                        0.2
This equation has one trivial solution φ = 0, next solution is slightly less than 3π/2 which
corresponds to the case H     > R (and is thus not correct). The one that gives a lower
bound for H is the third solution
                                           φ ≈ 5π/2
The equation φ = tan φ can be solved graphically or numerically to obtain a close value
(φ = 7.725 rad) which means

                                H/R = (1 − cos φ) ≈ 0.871
Give points if the method is correct, depending on the value of H/R found, according to
these intervals:                                                                               0.4
0.85 ≤ H/R ≤ 0.88: 0.4 pts
0.5 ≤ H/R < 0.85: 0.3 pts
0 < H/R < 0.5 or H > 0.88: 0.2 pts
H = 0 or method is incorrect: 0 pts

Solution two
relation between H and rotated angle φ

                                     R−H
                                         = cos φ                                               0.2
                                      R
obtain equation of the form
                                           p                !
                           H                 1 − (1 − H/R)2
                             = 1 − cos                                                         0.2
                           R                   1 − H/R
                                           √
                                              1−(1−x)2
Figure 3 gives a plot of f (x) = 1 − cos                 . The goal is to find an approximate
                                               1−x
solution for the second intersection. The first intersection is discarded  it is introduced
because of cos φ = cos(−φ) and corresponds to a situation with H > R.
Realizing that there is an infinite number of solutions.                                        0.2

                                                9

                          Figure 3:    Plot of f (H/R) and H/R

                              Figure 4:       Plot of g(x) and x

   • introduce new variable x := 1 − H/R, so that the equation becomes
                                        p
                                x = cos( 1 − x2 /x) =: g(x)

   • g(x) is then smaller than x up to the first solution. In particular it is negative in
     some region (see figure 4). Finding the third zero thus gives a lower bound for the
     solution:                              √
                                                1 − x2
                                                       = 5π/2
                                                  x
   • give lower bound
                       p                           p
                 x = 1/ 25π 2 /4 + 1 ⇒ H = R(1 − 1/ 25π 2 /4 + 1) ≈ 0.874

     Note: the actual result is H/R = 0.871 . . ..

     Use the same points for the numerical answer as was mentioned in solution one.             0.4
     If the student plots f rather than g , find solution to f       = 1: is equivalent to the
     solution above.   Give same number of points.
                                      √
                                           1−x2
     It is also possible to use cos
                                            x         = sin(1/x).

B8 (1.7 pt) Alice pulls the mass a distance d downwards from the equilibrium point
x = 0, y = 0, and then lets it go (see figure 4).

                                                      10

   • Give an algebraic expression of x(t) and y(t). You may assume that ωss d is small.

   • Sketch the trajectory (x(t), y(t)), marking all important features such as amplitude.

Solution B8:                                                                                [1.7]

Note:     we did not specify the overall sign of the Coriolis force.       Give same amount
of points if using opposite convention, but it has to be consistent! Otherwise: subtract
0.1pt for each instance of inconsistency.                                                     -0.1
Students are allowed to express everything in terms of   ω , they don't need to write
 k/m − ωss explicitly. Deduct 0.1pt however if they use k/m instead of ω ..                   -0.1
p
        2

Realize that y(t) is standard harmonic oscillation:

                                      y(t) = A cos ωt + B                                     0.1
Give correct constants from initial conditions

                                       y(t) = −d cos ωt                                       0.2
Correct expression for vy (t):
                                      vy (t) = −dω sin ωt                                     0.1
Coriolis force in x-direction

                         Fx (t) = 2mωss vy (t) = −2mωss dω sin ωt                             0.2
           Realize that this implies that x(t) is also a harmonic oscillation. . .            0.1
                . . . but with a constant movement term superimposed: vt                      0.1
getting the correct amplitude:
                                                 2ωss d
                                           A=                                                 0.1
                                                   ω
Correct answer with correct initial conditions:

                                          2ωss d
                                 x(t) =          sin ωt − 2ωss dt                             0.2
                                            ω
Sketch:

                             A
                                            y

                                            d

                                                                       x

                                            -d
                   B         4πωssd
                               ω
                         C

                                                  11

Correct qualitative sketch:
                                   periodic motion                        0.1
                              overall constant movement                   0.1
                                      B): cusps                           0.1
And additionally correct quantitative sketch:

                      A)+B): peaks and cusps are at y = ±d                0.1
                                                4πωss d
              C): cusps are at distance ∆x =            from each other   0.2
                                                  ω

                                           12

English (Official)                        Q1-1


[User] Veekahur tulistab veejuga, mille ristlõikepindala on $S = \SI{8}{cm^2}$ ning võimsus $N = \SI{6000}{W}$. Millise jõuga tabab veejuga märki, kui kahur ja märk asuvad samal kõrgusel? Vee tihedus on $\rho = \SI{1000}{kg/m^3}$, õhutakistust mitte arvestada. Märklaua ja veekahuri vahemaa on väike, st veejoa kõverdumist raskusjõu toimel ei pea arvestama.

Solution:
Olgu veejoa kiirus kahurist väljudes $v$. Et märk ja kahur asuvad samal kõrgusel ning õhutakistus puudub, siis märgini jõuab veejuga samuti kiirusega $v$. Aja $t$ jooksul väljub kahurist vesi massiga $m = \rho Svt$. Väljuva vee kineetiline energia on $E = mv^2/2$ ning asendades massi leiame
\[
E = \frac{\rho Sv^3t}{2}.
\]
Seega saame avaldada kahuri võimsuse
\[
N=\frac{E}{t}=\frac{\rho S v^{3}}{2} \quad \Rightarrow \quad v=\sqrt[3]{\frac{2 N}{\rho S}}.
\]
Jõud, millega veejuga tabab märki, on määratud vee impulsiga:
\[
F=\frac{m v}{t} \quad \Rightarrow \quad F=\frac{\rho S v^{2} t}{t}=\rho S v^{2}.
\]
Asendades $v$ saame
\[
F=\sqrt[3]{4 N^{2} \rho S} \approx \SI{490}{N}.
\]

Veekahur

Veekahur tulistab veejuga, mille ristlõikepindala on $S = \SI{8}{cm^2}$ ning võimsus $N = \SI{6000}{W}$. Millise jõuga tabab veejuga märki, kui kahur ja märk asuvad samal kõrgusel? Vee tihedus on $\rho = \SI{1000}{kg/m^3}$, õhutakistust mitte arvestada. Märklaua ja veekahuri vahemaa on väike, st veejoa kõverdumist raskusjõu toimel ei pea arvestama.

Õhutakistuse puudumisel jõuab veejuga märgini sama kiirusega nagu väljudes (energia jäävusest tulenevalt). Jõu leidmiseks võib vaadelda ajavahemikku $\Delta t$ ning selle jooksul üle antavat vee impulssi.


[User]
Kärbes on merevaigutükis, mille murdumisnäitaja on $n=\num{1,6}$. Tüki üks pinnaosa on sfääriline kõverusraadiusega $r = \SI{3}{mm}$. Kui vaadata kärbse pead läbi selle pinnaosa, siis näib pea asuvat kõveruskeskpunkti läbival sirgel $k = \SI{5}{mm}$ sügavusel merevaigus. Kui sügaval on kärbse pea tegelikult?

\emph{Märkus:} kasutada väikeste nurkade lähendust $\tan \alpha \approx \sin \alpha \approx \alpha$, kus $\alpha \gg 1$ on väike nurk mõõdetuna radiaanides.

Solution:
Olgu $O$ pinna kõverusraadiuse keskpunkt ning $K$ ja $A$ vastavalt kärbse näiline ja tegelik asukoht. Allpool toodud joonis kujutab kärbsest alguse saanud kahe kiire $AB$ ja $AC$ edasist käiku. Kärbse näiline asukoht $K$ vastab kärbsest alguse saanud murdunud kiirte pikenduste lõikepunktile (joonisel $KB$ ja $KC$).

Arvestades väikeste nurkade korral kehtivat lähendust $\tan \alpha \approx \sin \alpha \approx \alpha$, võime murdumisseaduse kirjutada kujul:
\[
\frac{\sin \beta}{\sin \alpha}=n \approx \frac{\beta}{\alpha}
\]
Kolmnurkade $\triangle OCB$, $\triangle ACB$ ja $\triangle KCB$ kaudu avaldame kaare $\widehat{BC}$
\[
\gamma r=\delta a=\varphi k=\widehat{BC}.
\]
Kolmnurkade $\triangle KAB$ ja $\triangle AOB$ kaudu avaldame nurga $\angle BOC$
\[
\alpha+\delta=\gamma \quad \text { ja } \quad \beta+\varphi=\gamma.
\]

\begin{center}
	\includegraphics[width=\linewidth]{2008-v2g-10-lah}
\end{center}

Saame
\[
\begin{array}{lll}{(\gamma-\alpha) a=\gamma r} & {\Rightarrow} & {\gamma(a-r)=\alpha a} \\ {(\gamma-\beta) k=\gamma r} & {\Rightarrow} & {\gamma(k-r)=\beta k.}\end{array}
\]

Paneme tähele, et need kaks võrrandit oleks saanud otse siinusteoreemist kolmnurkade $\triangle AOB$ ja $\triangle KOB$ arvestusega, et tänu nurkade $\varphi$, $\delta$ ja $\gamma$ väiksusele $|KB| \approx k$ ja $|AB| \approx a$.

Arvestades, et $\beta /\alpha \approx n$, saame
\[
\frac{a-r}{k-r}=\frac{a}{n k} \quad \Rightarrow \quad a=\frac{n r k}{n k-k+r}=\frac{\num{1,6} \cdot \num{3} \cdot \num{5}}{\num{1,6} \cdot \num{5}-\num{5}+\num{3}}=\SI{4}{mm}.
\]

Kärbes

Kärbes on merevaigutükis, mille murdumisnäitaja on $n=\num{1,6}$. Tüki üks pinnaosa on sfääriline kõverusraadiusega $r = \SI{3}{mm}$. Kui vaadata kärbse pead läbi selle pinnaosa, siis näib pea asuvat kõveruskeskpunkti läbival sirgel $k = \SI{5}{mm}$ sügavusel merevaigus. Kui sügaval on kärbse pea tegelikult?

\emph{Märkus:} kasutada väikeste nurkade lähendust $\tan \alpha \approx \sin \alpha \approx \alpha$, kus $\alpha \gg 1$ on väike nurk mõõdetuna radiaanides.

Kärbse näiline asukoht vastab kärbsest alguse saanud klaasist väljunud murdunud kiirte pikenduste lõikepunktile. Selle jaoks tasub vaadelda kahte kiirt: mõlemad saavad alguse kärbsest, aga üks väljub merevaigutükist pinnaga risti ja teine väikse nurga all. Edasi tuleb kiirte geomeetriat ettevaatlikult uurida ning rakendada Snelli seadust.


[User]
Consider two identical, thin, symmetric mirrors, with reflection and transmission
    coefficients r and t, placed a distance L apart, with air in between them and outside them. This
    system is called a Fabry–Perot interferometer. A wave with wavenumber k hits the apparatus; we
    want to find the reflection and transmission coefficients rnet and tnet of the entire system.

     (a) Draw all paths that the light could take to be reflected, and to be transmitted.

     (b) By applying the principle of superposition and summing a geometric series, show that

                                                  rt2 e2ikL                  t2 eikL
                                   rnet = r +                ,   tnet =                .
                                                1 − r2 e2ikL              1 − r2 e2ikL
          Note that your answers may differ by phases, depending on your conventions for rnet and tnet .

      (c) Show that all the light is transmitted for some special values of L, even if r ≈ 1. That is,
          nearly ideal mirrors can become perfectly transparent! This is called resonant transmission,
          and it occurs because the reflected waves perfectly destructively interfere.

    For the rest of the problem, assume that L takes one of the special values found in part (c).

      (d) Using energy conservation, recover the result of problem 22(b).

      (e) Suppose that a laser with power P has been fired at the interferometer for a long time. Then,
          at a certain moment, the laser is suddenly switched off. Find the total energy of the light
          that travels from the interferometer back towards the laser after the laser is switched off. For
          simplicity, suppose that |t| ≪ 1, and give your answer in terms of P , L, |t|, and c.

      (f) Estimate the duration of the light pulse that travels back towards the laser.

Solution:
(a) Light can be immediately reflected from the first mirror. It can also go in between
     the mirrors and be reflected any number of times before leaving through either mirror.

 (b) For a wave to be transmitted, it must be transmitted through one mirror, reflected 2n times,
     then travel a distance of L and get transmitted out the other mirror. For each intermediate
     reflection, its amplitude gets a factor of α = reikL . Then we have

                                                                           t2 eikL
                              tnet = t2 eikL 1 + α2 + α4 + . . . =

                                                                        1 − r2 e2ikL
     as desired. As for reflection, we have

                                                                               rt2 e2ikL
                         rnet = r + t2 eikL α + α3 + α5 + . . . = r +

                                                                             1 − r2 e2ikL

 (c) The fraction of the energy transmitted is

                                                              |t|2
                                        T = |tnet |2 =                   .
                                                         |1 − r2 e2ikL |

     This is maximized when r2 e2ikL is real and positive, so that it is equal to |r|2 , giving

                                                      |t|2
                                              T =            = 1.
                                                    1 − |r|2

     This is a striking result: you can put two nearly perfect mirrors next to each other, and light
     of the right color will still go right through. This is because the light that goes go through the
     first can bounce around inside many times, eventually completely canceling the zeroth order
     reflected wave. This phenomenon is called resonant transmission.

 (d) In this case, we have |tnet |2 = 1, so energy conservation implies rnet = 0, which means

                                    (t2 /r)r2 e2ikL       t2 |r|2 /r       t2 |r|2 /r
                             0=r+                   = r +            = r +            .
                                     1 − r2 e2ikL         1 − |r|2             |t|2

     This is equivalent to
                                                  r2     t2

                                                      =− 2
                                                 |r|    |t|
     so that the phases of r2 and t2 differ by π, so the phases of r and t differ by π/2.

      (e) We’ve established that, in the steady state, the total reflected wave has zero amplitude. Let
          ∆t = 2L/c be the time for one round trip within the interferometer. For the first ∆t, the total
          reflected wave is “missing” the direct reflected wave, so it has an amplitude of magnitude r.
          For the next ∆t, it is missing both the direct reflected wave and the one involving one trip
          within the interferometer, and so on.
          To make this more quantitative, note that |r| = reikL , so that

              rnet = r + t2 eikL (|r| + |r|3 + |r|5 + . . .) = e−ikL |r| + (|r|3 − |r|) + (|r|5 − |r|3 ) + . . . .

          So for the first ∆t, the reflected amplitude has magnitude |r|, and for the next ∆t, it has
          magnitude |r|3 , and so on. Then we have

                                                                           2P L |r|2     PL
                             Er = (P ∆t)(|r|2 + |r|6 + |r|10 + . . .) =              4
                                                                                ≈ 2
                                                                            c 1 − |r|   |t| c
          where we used |t| ≪ 1.

      (f) Every time ∆t, the reflected pulse weakens by a factor of |r|4 , where 1 − |r|4 ≈ 2|t|2 . Then
          the timescale of decay is roughly T ∼ ∆t/|t|2 ∼ L/(|t|2 c).

                     01W
    This question is essentially the same as EuPhO 2024, problem 3.

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


[User]
7. Telescope (12 points)

As it is well known, a telescope makes it possib-
le to see the stars in daylight. Let us study the
problem in more details. Consider a simplified
model of the eye: a single lens with focal length
f = 4cm and diameter d = 3mm creating an
image onscreen (retina). Themodel ofa telesco-
pe is similar: a lens of focal length F = 2m and
diameter D = 20cm creating an image in focal
plane(whereeg.afilmcanbeput). Inyourcalcu-
lations, the following quantities can be used: the
density of the light energy radiated from a unit
Solar surface in unit time w0 (the light power
surface density); the ratio of the star and Sun
distances q = 4 · 105
(we assume that the star
is identical to the Sun); Solar angular diameter
φ ≈ 9mrad. Remark: If the answer contains w0
then numerical answeris not required.
1)Considerasheetofpaper,thenormalofwhich
is directed towards the Sun. What is the sur-
face density of the light power w1 arrivingto the
sheet fromthe Sun?
2) Find the net power P2 of the light, which is
focused by the telescope into the image of the
star.
3) Assume that blue sky is as bright as a sheet
of gray paper illuminated by Sun. You may as-
sume that in the direction, perpendicular to the
sheet, the ratio of the light power scattered by
the paper into a 1-steradian space angle, to the
net light power arriving to the sheet, is α ≈ 0,1
(this corresponds to the dissipation of ca 70 %
light energy in the gray paper). What is the sur-
face density of the light power in the focal plane
of the telescope w3, due to the blue sky?
4) While studying the star image, let us ignore
all the effects other than diffraction. Estimate
the surface density of the light power in the cen-
ter of the star image w2 (in the focal plane of the
telescope),dueto the light arrivingfromthe star.
5) Provide an expression for the ratio of the sur-
face densities of the light powers k in the midd-
le of the star image, and in a point farther away
from it.
6) Is it possible to see a star in daylight using a
telescope? Plain eye? Motivate yourself.

Solution:
7. Telescope

1) The light flux density decreases inversely pro-
portionally to the square of the distance, the-
refore w1 = w0R2
p/L2
p, where Rp is the solar
radius, and Lp — the solar distance. Due to
φ = 2Rp/Lp, we obtain w1 = w0φ2
/4.
2) The previous result can be applied to the
star flux density, which is q−2
w1; hence P2 =
1
4πD2
w1q−2
= w0π(φD/4q)2
.
3) The paper surface area S radiates towards
the lens of the telescope the power P3 =
w1αS(π
4 D2
/L2
), where L is the telescope dis-
tance. The image of this piece of paper has
size s = SF2
/L2
; thus, w3 = P3/s =
w1α(π
4 D2
/F2
) = w0απ(φD/4F)2
.
4) The angular distance of the first diffraction
minimum (using the single slit approximation
— circle is actually not a slit) is λ/D. Hence,
the bright circle radius can be estimated as
δ = Fλ/D. Consequently, w2 = P2/πδ2
=
w0(φD2
/4qFλ)2
.
5)k = (w2 +w3)/w3 = 1+(απ)−1
(D/λq)2
≈
4 (assuming λ ≈ 500nm).
6) k − 1 ∼ 1 (or k − 1 > 1) means that the star
can be easily seen (as is the case for the telesco-
pe); k − 1  1 means that the star cannot be
seen (for the eye,k − 1 ≈ 1 · 10−4
).

Telescope


[User]
Kui suure kiirendusega $a_k$ ja mis suunas hakkab liikuma kahest kehast koosneva süsteemi masskese, kui kehad on seotud niidiga, mis on tõmmatud üle ploki (vt joonist)? Kehade massid on $m_1$ ja $m_2$ ($m_1$ < $m_2$), niit on kaalutu ja mitteelastne.
\begin{center}
	\includegraphics[width=0.25\linewidth]{2006-lahg-05-yl}
\end{center}

Solution:
Kuna raskem keha hakkab liikuma allapoole ja kergem ülespoole, siis on selge, et süsteemi massikese hakkab liikuma allapoole.

Olgu süsteemi massikese alghetkel punktis $C_0$ (vt joonist). Kehade massikeskmete kaugused süsteemi massikeskmest leiame tingimusest:
\begin{equation} \label{2006-lahg-05:eq1}
m_1b_1 = m_2b_2.
\end{equation}
Aja $t$ jooksul liigub raskem keha allapoole ja kergem liigub ülespoole kauguse $H$ võrra:

\begin{center}
	\includegraphics[width=0.4\linewidth]{2006-lahg-05-lah}
\end{center}

\begin{equation} \label{2006-lahg-05:eq2}
H = at^2 2,
\end{equation}
kus $a$ on kehade kiirendus. Süsteemi masskese liigub sama aja jooksul kauguse $h$ võrra, mille määrab tingimus:
\[
m_2 (H + b_2 - h) = m_1 (H + b_1 + h).
\]
Siit, arvestades valemeid (\ref{2006-lahg-05:eq1}) ja (\ref{2006-lahg-05:eq2}), leiame, et:
\begin{equation} \label{2006-lahg-05:eq3}
h=\frac{m_{2}-m_{1}}{m_{2}+m_{1}} H=\frac{m_{2}-m_{1}}{m_{2}+m_{1}} \frac{a t^{2}}{2}.
\end{equation}
Kiirenduse $a$ leiame võrrandisüsteemist:
\[
m_2a = m_2g - T,\quad m_1a = T - m_1g,
\]
kus $T$ on niidi tõmbepinge. Avaldades $a$ võrrandisüsteemist, saame
\[
a=\frac{m_{2}-m_{1}}{m_{2}+m_{1}} g.
\]
Asendades leitud väärtuse valemisse (\ref{2006-lahg-05:eq3}), saame:
\[
\begin{aligned}
h&=\left(\frac{m_{2}-m_{1}}{m_{2}+m_{1}}\right)^{2} \frac{g t^{2}}{2}=\frac{a_{k} t^{2}}{2} \quad \Rightarrow \\
&\Rightarrow \quad a_{k}=\left(\frac{m_{2}-m_{1}}{m_{2}+m_{1}}\right)^{2} g.
\end{aligned}
\]

Plokk

Kui suure kiirendusega $a_k$ ja mis suunas hakkab liikuma kahest kehast koosneva süsteemi masskese, kui kehad on seotud niidiga, mis on tõmmatud üle ploki (vt joonist)? Kehade massid on $m_1$ ja $m_2$ ($m_1$ < $m_2$), niit on kaalutu ja mitteelastne.
\begin{center}
	\includegraphics[width=0.25\linewidth]{2006-lahg-05-yl}
\end{center}

Ülesandes ei ole niidi pinge $T$ teada, aga see on leitav pannes plokkide jaoks kirja Newtoni teise seaduse ning niidi venimatuse tingimuse. Massikeskme kiirenduse avaldamine plokkide kiirenduste kaudu on analoogne massikeskme koordinaadi avaldamisega plokkide koordinaatide kaudu.


[User]
Muudetava mahtuvusega kondensaator on ühendatud patareiga,
mille klemmide pinge on $U$. Kondensaatori mahtuvust muudetakse laadimisel nii, et
kondensaatori laadimise vool $I$ on konstantne. Leidke patarei võimsus ja kondensaatori laadimisel energia salvestamise kiirus.
Põhjendage võimalikku erinevust.

Solution:
Patarei pinge on $U=\text{const}$ ja vool ahelas, vastavalt ülesande tingimustele, on $I=\text{const}$. Patarei võimsus on
\[P_p=UI. \]
Energia kondensaatoris on
\[E=\frac{CU^2}{2}, \]
kus kondensaatori mahtuvus on $C=q/U$ ja laeng kondensaatoris $q$.
Kondensaatorisse energia salvestamise kiirus on energia muutumise kiirus kondensaatoris ehk energia tuletis aja järgi,
\[P_C=\frac{\D E}{\D t}=\frac{\D C}{\D t}\frac{U^2}{2}=\frac{\D (CU)}{\D t}\frac{U}{2}=\frac{\D q}{\D t}\frac{U}{2}.\]
Arvestades, et laengu muutumise kiirus $\D q/\D t$ on vool $I$, saame energia salvestumise kiiruseks kondensaatorisse
\[P_C=\frac{UI}{2}.\]
Näeme, et patareist \enquote{väljub} energiat kaks korda kiiremini, kui seda salvestub kondensaatorisse. Energia, mis ajaühikus kondensaatorisse ei jõua, läheb välisjõududele, mis muudavad kondensaatori mahtuvust selliselt, et $I=\text{const}$, vastu töö tegemiseks.

Kondensaator

Muudetava mahtuvusega kondensaator on ühendatud patareiga,
mille klemmide pinge on $U$. Kondensaatori mahtuvust muudetakse laadimisel nii, et
kondensaatori laadimise vool $I$ on konstantne. Leidke patarei võimsus ja kondensaatori laadimisel energia salvestamise kiirus.
Põhjendage võimalikku erinevust.

Patarei tehtud töö ajaühikus $\Delta t$ on $UI\Delta t$. Kondensaatorisse salvestatava energia muutumise kiirus on leitav võttes kondensaatori siseenergiast $\frac{CU^2}{2}$ aja järgi tuletise.


[User]
1. Asteroid (14 points)

Consider a hypothetical asteroid of mass ma and radius
ra, which moves along an elliptical orbit around the Sun (of
mass Ms) in the same direction as Earth. Let us assume that
the Earth’s orbit is a circle of radius Re (neglecting thus its
eccentricity) and that the two orbits lay in the same plane.
The asteroid’s shortest distance to the Sun (at its perihe-
lion) Rmin = 1
2Re and the longest distance (at the apohelion)
Rmax = 1.51Re ≈ 1.5Re (you may use the approximate value
to simplify your calculations). The orbital velocity of the Earth
v0 = 30km/s. You can also use the following numerical val-
ues: the radius of the Earth re = 6400km, free fall acceler-
ation at the Earth’s surface g = 9.81m/s2
, angular diameter
of the Sun as seen from the Earth α = 0.5◦
, duration of one
year T0 = 365 days, the temperature of the Sun’s surface Ts =
6000K, free fall acceleration at the Sun’s surface gs = 275m/s2
,
Stefan Boltzmann constant σ = 5.6704× 10−8
kg·s−3
·K−4
,
the speed of light c = 3 × 108
m/s. The asteroid is of a spher-
ical shape, its radius ra = 10m and mass ma = 1 × 107
kg;
both the Sun and the asteroid can be considered as perfectly
black bodies.
Part A. Collision with Earth (5 points)
i. (2 pts) Suppose that the asteroid will collide with the Earth
and is already very close, at a distance l ≪ Re from the Earth’s
surface; what is the velocity of the asteroid with respect to the
Earth assuming that (a) l ≫ re; (b) l ≪ re.
ii. (2 pts) Impact parameter b is defined in the Earth’s refer-
ence frame as the distance between the Earth and a line, tan-
gent to the asteroid’s trajectory at a point which is far enough
(at a distance l, re ≪ l ≪ Re). Determine the maximal value
of the impact parameter bmax for which the asteroid will still
collide with the Earth.
iii. (1 pt) According to calculations, the asteroid is going to
hit the Earth centrally after N = 10 orbital periods. In order
to avert the collision, the period of the asteroid needs to be
changed; by how many seconds? (Assume simplifyingly that
the intersection point of the two orbits remains at rest.)
Part B. Changing the solar pull (9 points)
Theoretically, it is possible to change the asteroid’s period
by making use of the pressure of the solar radiation. Let us
study, how realistic is such a project. The Sun pulls the as-
teroid with the gravitational force equal to F0 = GMsma/R2
,
where G is the gravitational constant and R is the distance
between the Sun and the asteroid at the given moment of time.
Let us denote GMs = γ0, so that
F = γ0ma/R2
.
Suppose that when the asteroid is at its perihelion, the con-
stant γ0 is decreased instantaneously, down to a new value γ1,
which remains constant during the subsequent motion.
i. (2 pts) Find the new apohelion distance R′
max of the aster-
oid; express it in terms of κ = (γ0 − γ1)/γ0.
ii. (2 pts) Find the change of the asteroid’s orbital period
assuming that κ ≪ 1 (express it in terms of κ).
iii. (4 pts) Suppose that at its perihelion, the asteroid is coated
with a perfectly retro-reflective paint (which directs all the in-
cident light directly back, towards the source). Such a painting
will result in a change of the effective pull of the asteroid to-
wards the Sun; find the respective value of κ (provide also a
numerical estimate).
iv. (1 pt) Estimate, is it realistic to avert the collision with
this asteroid using the retroreflective paint.
Problem

Solution:
1. Asteroid

Part A. Collision with Earth (5 points)
i. (2 pts) The longer axis of the asteroid 2a = Rmax +Rmin =
2Re equals to that of Earth, so the full energies, when reduced
to the unit mass, are equal. Immediately before the collision,
the Earth and the asteroid are at the same distance from the
Sun, so the gravitational potentials are equal, too. Hence, the
speeds are also equal. The distance between the Sun and the
asteroid equals to the longer semiaxis, hence it is situated at
the shorter semiaxis of the orbit. The velocity of the asteroid
is perpendicular to the shorter semiaxis, and the velocity of the
Earth — to the radius vector drawn from the Sun. So, the angle
between those two vectors is the angle between the radius vec-
tor and the shorter semiaxis, sinα = 1
2(Rmax −Rmin)/Re = 1
2,
hence α = 30◦
. The relative velocity of the asteroid is the
vector difference of the two vectors, so its modulus equals to
va = 2v0 sin15◦
≈ 15.5km/s. When accelerated further by the
Earth’s gravity field, the respective gravitational energy will be
added to the kinetic one, vb =
p
v2
a + 2gre = 19.1km/s.
ii. (2 pts) At the limit case of impact, the trajectory of the
asteroid is tangent to the surface of the Earth. So, we can
apply the conservation of angular momentum for the point
where the trajectory touches the Earth, vab = vbre, hence
b = revb/va = 7900km.
iii. (1 pt) Suppose that the asteroid is delayed by τ; at that
moment when the asteroid is at the Earth’s orbit, the Earth
is at the distance l = v0τ from the asteroid. The relative ve-
locity of the asteroid forms with this displacement vector an
angle equal to 90◦
− 15◦
= 75◦
, hence the impact parameter
b = v0τ sin75◦
, from where τ = b/v0 sin75◦
≈ 270s. Since this
time delay is accumulated over 10 periods, the delay need for
a single period is τ/10 = 27s.
Part B. Changing the solar pull (9 points)
i. (2 pts) When γ changes, the kinetic energy remains con-
stant:
−
γ0
2a
+
γ0
0.5Re
= −
γ1
2a′
+
γ1
0.5Re
where a = Re and 2a′
= 0.5Re + R′
max. So,
γ1
2a′
=
γ0
2Re

4
γ1
γ0
− 3

⇒ (1 − κ)Re = a′
(1 − 4κ),
a′
= Re
1 − κ
1 − 4κ
, R′
max =
Re
2
3
1 − 4κ
.
ii. (2 pts) At the limit of small κ, we can simplify the previous
result,
a′
Re
≈ 1 + 3κ.
From the Kepler’s third law, T/T0 = (a′
/a)3/2
p
γ0/γ1, from
where ∆T
T0
≈ 3
2
a′
−a
a + κ
2 = 5κ. So, ∆T = 5T0κ.
iii. (4 pts) For photons, the energy-to-mass ratio is c. There-
fore, at the Sun’s surface, the momentum carried by photons
per unit time across a surface area S is given by dp/dt =
SσT4
s /c. As the result of the coating, the photons are re-
flected back by the asteroid, instead of being absorbed. So,
before coating, each photon gave to the asteroid a momentum
equal to its own; no it will double. Hence, the change in the
force due to photons is given by ∆F = πr2
aσT4
s /c (assuming
that the asteroid is at the Sun’s surface). Both the pressure
of photons and gravity force are inversely proportional to the
distance from the Sun, so the force due to photons can be, in-
deed, considered as a correction to the gravity constant. κ is
the relative change of that constant and can be calculated for
the Sun’s surface as
κ = ∆F/gSma = πr2
aσT4
s /cgSma ≈ 2.8 × 10−8
.
iv. (1 pt) We need to have ∆T = 27s, hence κ = 1
5
∆T
T0
≈
1.7 × 10−7
. This exceeds by an order of magnitude the effect
provided by the coating. κ provided by the coating is inversely
proportional to the diameter of the asteroid; the required κ is
inversely proportional to N. So, it would be possible to avert
collision for ra = 2m, or for ra = 10m with N = 60. In the
first case, the asteroid may not be large enough to warrant at-
tention; in the second case, 60 years is too long time. So, the
answer is “no”.
Problem

Asteroid