IPhO 2016 · Theory problem 1

Two Problems in Mechanics

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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 \(r_1\) and thickness \(h_1\). Somewhere inside the wooden cylinder, the wood has been replaced by a metal disk of radius \(r_2\) and thickness \(h_2\). 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 \(\rho_1\), the density of the metal is \(\rho_2>\rho_1\). 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:

\[ r_1,\ h_1,\ \rho_1,\ \rho_2,\ M. \tag{1} \]

The goal is to determine \(r_2\), \(h_2\) and \(d\), through indirect measurements.

Figure 1

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 \(\Theta\) (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 \(\phi\) which we measure.

Figure 2

Figure 2: Cylinder on an inclined base.

A.1 — 0.8pt

Find an expression for \(b\) as a function of the quantities (1), the angle \(\phi\) and the tilting angle \(\Theta\) of the base.

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

Figure 3

Figure 3: Suspended system.

Next we want to measure the moment of inertia \(I_S\) 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 \(\varphi\), and let it go. See figure 3 for the setup. We find that \(\varphi\) describes a periodic motion with period \(T\).

A.2 — 0.5pt

Find the equation of motion for \(\varphi\). Express the moment of inertia \(I_S\) of the 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 \(\varphi\) 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 — 0.4pt

Find an expression for the distance \(d\) as a function of \(b\) and the quantities (1). You may also include \(r_2\) and \(h_2\) as variables in your expression, as they will be calculated in subtask A.5.

A.4 — 0.7pt

Find an expression for the moment of inertia \(I_S\) in terms of \(b\) and the known quantities (1). You may also include \(r_2\) and \(h_2\) as variables in your expression, as they will be calculated in subtask A.5.

A.5 — 1.1pt

Using all the above results, write down an expression for \(h_2\) and \(r_2\) in terms of \(b\), \(T\) and the known quantities (1). You may express \(h_2\) as a function of \(r_2\).

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 — 0.5pt

At what angular frequency \(\omega_{ss}\) does the space station rotate so that the astronauts experience the same gravity \(g_E\) 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 — 0.2pt

Assuming that on Earth gravity is constant with acceleration \(g_E\), what would be the angular oscillation frequency \(\omega_E\) that a person on Earth would measure?

B.3 — 0.6pt

What angular oscillation frequency \(\omega\) does Alice measure on the space station?

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.

Figure 4

Figure 4: Space station

B.4 — 0.8pt

Derive an expression of the gravity \(g_E(h)\) for small heights \(h\) above the surface of the Earth and compute the oscillation frequency \(\widetilde{\omega}_E\) of the oscillating mass (linear approximation is enough). Denote the radius of the Earth by \(R_E\). 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 — 0.3pt

For what radius \(R\) of the space station does the oscillation frequency \(\omega\) match the oscillation frequency \(\widetilde{\omega}_E\) on the Earth? Express your answer in terms of \(R_E\).

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 \(\vec F_C\) called the Coriolis force. The force \(\vec F_C\) acting on an object of mass \(m\) moving at velocity \(\vec v\) in a rotating frame with constant angular frequency \(\vec\omega_{ss}\) is given by

\[ \vec F_C=2m\vec v\times\vec\omega_{ss}. \tag{2} \]

In terms of the scalar quantities you may use

\[ F_C=2mv\omega_{ss}\sin\phi, \tag{3} \]

where \(\phi\) 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.

B.6 — 1.1pt

Calculate the horizontal velocity \(v_x\) and the horizontal displacement \(d_x\) (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 \(d_x\ll H\).

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 \(d_x=0\).

B.7 — 1.3pt

Find a lower bound for the height of the tower for which it can happen that \(d_x=0\).

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.

Figure 5

Figure 5: Setup.

B.8 — 1.7pt

Alice pulls the mass a distance \(d\) downwards from the equilibrium point \(x=0\), \(y=0\), and then lets it go (see figure 5).

  • Give an algebraic expression of \(x(t)\) and \(y(t)\). You may assume that \(\omega_{ss}d\) 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.

Alice and Bob continue to argue.

General Data Sheet

Speed of light in vacuum \(c\) \(=299\,792\,458\ \mathrm{m\cdot s^{-1}}\)
Vacuum permeability (magnetic constant) \(\mu_0\) \(=4\pi\times10^{-7}\ \mathrm{kg\cdot m\cdot A^{-2}\cdot s^{-2}}\)
Vacuum permittivity (electrical constant) \(\varepsilon_0\) \(=8.854\,187\,817\times10^{-12}\ \mathrm{A^2\cdot s^4\cdot kg^{-1}\cdot m^{-3}}\)
Elementary charge \(e\) \(=1.602\,176\,620\,8(98)\times10^{-19}\ \mathrm{A\cdot s}\)
Mass of the electron \(m_{\mathrm e}\) \(=9.109\,383\,56(11)\times10^{-31}\ \mathrm{kg}\)
\(=0.510\,998\,946\,1(31)\ \dfrac{\mathrm{MeV}}{c^2}\)
Mass of the proton \(m_{\mathrm p}\) \(=1.672\,621\,898(21)\times10^{-27}\ \mathrm{kg}\)
\(=938.272\,081\,3(58)\ \dfrac{\mathrm{MeV}}{c^2}\)
Mass of the neutron \(m_{\mathrm n}\) \(=1.674\,927\,471(21)\times10^{-27}\ \mathrm{kg}\)
\(=939.565\,413\,3(58)\ \dfrac{\mathrm{MeV}}{c^2}\)
Unified atomic mass unit \(u\) \(=1.660\,539\,040(20)\times10^{-27}\ \mathrm{kg}\)
Rydberg constant \(R_\infty\) \(=10\,973\,731.568\,508(65)\ \mathrm{m^{-1}}\)
Universal constant of gravitation \(G\) \(=6.674\,08(31)\times10^{-11}\ \mathrm{m^3\cdot kg^{-1}\cdot s^{-2}}\)
Acceleration due to gravity (in Zurich) \(g\) \(=9.81\ \mathrm{m\cdot s^{-2}}\)
Planck’s constant \(h\) \(=6.626\,070\,040(81)\times10^{-34}\ \mathrm{kg\cdot m^2\cdot s^{-1}}\)
Avogadro number \(N_A\) \(=6.022\,140\,857(74)\times10^{23}\ \mathrm{mol^{-1}}\)
Molar gas constant \(R\) \(=8.314\,4598(48)\ \mathrm{kg\cdot m^2\cdot s^{-2}\cdot mol^{-1}\cdot K^{-1}}\)
Molar mass constant \(M_u\) \(=1\times10^{-3}\ \mathrm{kg\cdot mol^{-1}}\)
Boltzmann constant \(k_B\) \(=1.380\,648\,52(79)\times10^{-23}\ \mathrm{kg\cdot m^2\cdot s^{-2}\cdot K^{-1}}\)
Stefan-Boltzmann constant \(\sigma\) \(=5.670\,367(13)\times10^{-8}\ \mathrm{kg\cdot s^{-3}\cdot K^{-4}}\)

Official English text of the problem.

Answers

For Part A, define the mass of an entirely wooden cylinder and the excess mass introduced by the metal: \[ m_0=\pi\rho_1r_1^2h_1, \qquad \Delta m=M-m_0 =\pi(\rho_2-\rho_1)r_2^2h_2. \] For Part B, write \(\Omega=\omega_{ss}\), and choose \(+x\) along the station’s rotation and \(+y\) towards its centre.

  • A.1 \[ \boxed{b=\frac{r_1\sin\Theta}{\sin\phi}}. \]

  • A.2 \[ \boxed{I_S\ddot\varphi=-Mgb\sin\varphi\simeq-Mgb\varphi}, \qquad \boxed{I_S=\frac{MgbT^2}{4\pi^2}}. \]

  • A.3 \[ \boxed{d=\frac{Mb}{\Delta m} =\frac{Mb}{\pi(\rho_2-\rho_1)r_2^2h_2}}. \]

  • A.4 \[ \boxed{I_S=\frac12m_0r_1^2+\frac12\Delta m\,r_2^2 +\frac{M^2b^2}{\Delta m}}. \]

  • A.5 \[ \boxed{ r_2=\sqrt{ \frac{MgbT^2}{2\pi^2\Delta m} -\frac{m_0r_1^2}{\Delta m} -\frac{2M^2b^2}{(\Delta m)^2}}}, \qquad \boxed{h_2=\frac{\Delta m}{\pi(\rho_2-\rho_1)r_2^2}}. \]

  • B.1 \[ \boxed{\omega_{ss}=\sqrt{\frac{g_E}{R}}}. \]

  • B.2 \[ \boxed{\omega_E=\sqrt{\frac{k}{m}}}. \]

  • B.3 \[ \boxed{\omega=\sqrt{\frac{k}{m}-\Omega^2} =\sqrt{\frac{k}{m}-\frac{g_E}{R}}}. \]

  • B.4 \[ \boxed{g_E(h)=\frac{g_E}{(1+h/R_E)^2} \simeq g_E\left(1-\frac{2h}{R_E}\right)}, \qquad \boxed{\widetilde{\omega}_E =\sqrt{\frac{k}{m}-\frac{2g_E}{R_E}}}. \]

  • B.5 \[ \boxed{R=\frac{R_E}{2}}. \]

  • B.6 \[ \boxed{v_x=-2\Omega H}, \qquad \boxed{d_x=-\frac23\Omega H\sqrt{\frac{2H}{g_E}} =-\frac{2\sqrt2}{3}H\sqrt{\frac{H}{R}}}. \] Both are opposite to the station’s rotation.

  • B.7 A rigorous lower bound is \[ \boxed{H>R\left(1-\frac{1}{\sqrt{1+4\pi^2}}\right) \simeq0.8428R}. \] The first exact return to the tower’s base occurs at \[ \boxed{H_{\min}\simeq0.8716R}. \]

  • B.8 With \(a=2\Omega d/\omega\), \[ \boxed{x(t)=a\bigl[\omega t-\sin(\omega t)\bigr]}, \qquad \boxed{y(t)=-d\cos(\omega t)}. \] The trajectory consists of successive cycloidal arches, with cusps at \((2\pi n a,-d)\), upper turning points at \(((2n+1)\pi a,d)\), and horizontal spacing \(2\pi a\). The vertical amplitude is \(d\).

Solution

A.1

Let \(P\) be the contact point with the inclined base. The normal force and static friction both act at \(P\), so they exert no torque about \(P\). Static equilibrium therefore requires the vertical line through the centre of mass \(C\) to pass through \(P\).

The centre of mass lies along \(SB\). Using the angles shown in Figure 2, the horizontal displacements of \(P\) and \(C\) from \(S\) are respectively \[ x_P-x_S=r_1\sin\Theta, \qquad x_C-x_S=b\sin\phi. \] Consequently, \[ Mg\left(b\sin\phi-r_1\sin\Theta\right)=0, \] and hence \[ \boxed{b=\frac{r_1\sin\Theta}{\sin\phi}}. \]

The stable equilibrium is the branch with \(\cos\phi>0\). In particular, the inclination must be small enough that \(r_1\sin\Theta<b\); equality corresponds to the limiting, non-restoring configuration.

A.2

The suspension fixes the axis \(S\). Gravity acts at \(C\), a distance \(b\) from this axis, and produces the restoring torque \[ \tau_S=-Mgb\sin\varphi. \] The rotational equation of motion is therefore \[ I_S\ddot\varphi=-Mgb\sin\varphi. \]

For \(|\varphi|\ll1\), use \(\sin\varphi\simeq\varphi\): \[ \ddot\varphi+\frac{Mgb}{I_S}\varphi=0. \] This is simple harmonic motion with \[ \left(\frac{2\pi}{T}\right)^2=\frac{Mgb}{I_S}. \] Thus \[ \boxed{I_S=\frac{MgbT^2}{4\pi^2}}, \] where the supplied gravitational acceleration is \(g=9.81\ \mathrm{m\,s^{-2}}\).

A.3

Represent the actual density distribution as the sum of:

  1. A complete wooden cylinder, of mass \[ m_0=\pi\rho_1r_1^2h_1, \] centred on \(S\).
  2. An additional disk of density \(\rho_2-\rho_1\), centred on \(B\), with mass \[ \Delta m=\pi(\rho_2-\rho_1)r_2^2h_2=M-m_0. \]

This accounts for the wood that was removed before the metal was inserted.

Taking the first moment of mass about \(S\), \[ Mb=m_0(0)+\Delta m\,d. \] It follows that \[ \boxed{ d=\frac{Mb}{\Delta m} =\frac{Mb}{M-\pi\rho_1r_1^2h_1} =\frac{Mb}{\pi(\rho_2-\rho_1)r_2^2h_2}}. \]

A.4

The complete wooden cylinder contributes \[ I_{\mathrm{wood}}=\frac12m_0r_1^2. \] The additional disk has moment of inertia \(\frac12\Delta m\,r_2^2\) about its own symmetry axis. By the parallel-axis theorem, its contribution about \(S\) is \[ I_{\mathrm{additional}} =\frac12\Delta m\,r_2^2+\Delta m\,d^2. \] Therefore \[ I_S=\frac12m_0r_1^2 +\frac12\Delta m\,r_2^2+\Delta m\,d^2. \]

Substituting \(d=Mb/\Delta m\) gives \[ \boxed{ I_S=\frac12m_0r_1^2 +\frac12\Delta m\,r_2^2 +\frac{M^2b^2}{\Delta m}}. \]

A.5

Equating the expressions for \(I_S\) from A.2 and A.4, \[ \frac{MgbT^2}{4\pi^2} =\frac12m_0r_1^2+\frac12\Delta m\,r_2^2 +\frac{M^2b^2}{\Delta m}. \] Solving for \(r_2^2\), \[ r_2^2= \frac{MgbT^2}{2\pi^2\Delta m} -\frac{m_0r_1^2}{\Delta m} -\frac{2M^2b^2}{(\Delta m)^2}. \] Taking the positive root, \[ \boxed{ r_2=\sqrt{ \frac{MgbT^2}{2\pi^2\Delta m} -\frac{m_0r_1^2}{\Delta m} -\frac{2M^2b^2}{(\Delta m)^2}}}. \]

The mass relation then determines the thickness: \[ \boxed{ h_2=\frac{\Delta m}{\pi(\rho_2-\rho_1)r_2^2}}. \]

Here \(m_0=\pi\rho_1r_1^2h_1\) and \(\Delta m=M-m_0\), so all quantities on the right are known or measured. Physically consistent measurements must give a positive radicand and satisfy \(d+r_2\le r_1\) and \(h_2\le h_1\).

B.1

An astronaut at rest relative to the rim follows a circle of radius \(R\). The required centripetal acceleration is \[ a_c=\omega_{ss}^2R. \] The normal force supplying this acceleration is experienced as the astronaut’s weight. Matching the acceleration on Earth requires \[ \omega_{ss}^2R=g_E. \] Thus \[ \boxed{\omega_{ss}=\sqrt{\frac{g_E}{R}}}. \]

Write \(\Omega=\omega_{ss}\) below.

B.2

Let \(y\) denote upward displacement from the spring’s equilibrium position. Constant gravity changes the equilibrium extension, but its constant force cancels when the equation is written about equilibrium: \[ m\ddot y=-ky. \] Therefore \[ \boxed{\omega_E=\sqrt{\frac{k}{m}}}. \]

B.3

Use the rotating frame and let \(y\) point radially inward from equilibrium. If the equilibrium radius is \(r_0\), the instantaneous radius is \(r_0-y\).

The centrifugal force acts outward and has magnitude \[ m\Omega^2(r_0-y). \] The inward spring force decreases by \(ky\) when the mass moves inward by \(y\). At equilibrium, its value \(F_0\) satisfies \[ F_0=m\Omega^2r_0. \] The inward equation of motion is therefore \[ m\ddot y=(F_0-ky)-m\Omega^2(r_0-y) =-(k-m\Omega^2)y. \]

The horizontal constraint balances the Coriolis force, so it does not alter this radial equation. Hence \[ \boxed{ \omega=\sqrt{\frac{k}{m}-\Omega^2} =\sqrt{\frac{k}{m}-\frac{g_E}{R}}}. \]

Oscillations require \(k/m>\Omega^2\).

B.4

Let \(M_E\) be Earth’s mass. Newton’s gravitational law gives \[ g_E(h)=\frac{GM_E}{(R_E+h)^2}. \] Since \(g_E=GM_E/R_E^2\), \[ g_E(h)=g_E\left(1+\frac{h}{R_E}\right)^{-2}. \] For \(h\ll R_E\), the first-order expansion is \[ \boxed{ g_E(h)\simeq g_E\left(1-\frac{2h}{R_E}\right)}. \]

Let the equilibrium height be \(h_0\), and let \(y\) be upward displacement from equilibrium. After subtracting the equilibrium force balance, \[ m\ddot y =-ky-m\bigl[g_E(h_0+y)-g_E(h_0)\bigr]. \] Within the linear approximation, \[ g_E(h_0+y)-g_E(h_0)\simeq-\frac{2g_E}{R_E}y. \] Thus \[ m\ddot y=-\left(k-\frac{2mg_E}{R_E}\right)y, \] giving \[ \boxed{ \widetilde{\omega}_E =\sqrt{\frac{k}{m}-\frac{2g_E}{R_E}}}. \]

Gravity decreases with height, so an upward displacement reduces the downward gravitational force. This weakens the net restoring force and lowers the frequency.

B.5

Equating the squared frequencies from B.3 and B.4, \[ \frac{k}{m}-\frac{g_E}{R} =\frac{k}{m}-\frac{2g_E}{R_E}. \] Therefore \[ \boxed{R=\frac{R_E}{2}}. \]

B.6

Choose \(+x\) along the station’s rotation and \(+y\) inward, towards its centre. With this convention, the Coriolis acceleration components are \[ a_{C,x}=2\Omega\dot y, \qquad a_{C,y}=-2\Omega\dot x. \]

For a short tower, approximate the downward acceleration by the constant \(g_E=\Omega^2R\). The small horizontal deflection permits us to neglect its feedback on the vertical fall. Released from rest, \[ y(t)=H-\frac12g_Et^2, \qquad \dot y=-g_Et. \] The impact time is \[ t_f=\sqrt{\frac{2H}{g_E}}. \]

The horizontal equation becomes \[ \ddot x=2\Omega\dot y=-2\Omega g_Et. \] Using \(x(0)=\dot x(0)=0\), integration gives \[ \dot x=-\Omega g_Et^2, \qquad x=-\frac13\Omega g_Et^3. \]

At impact, \[ \boxed{v_x=-\Omega g_Et_f^2=-2\Omega H}, \] and \[ \boxed{ d_x=-\frac13\Omega g_Et_f^3 =-\frac23\Omega H\sqrt{\frac{2H}{g_E}}}. \] Using \(\Omega^2=g_E/R\), \[ \boxed{ d_x=-\frac{2\sqrt2}{3}H\sqrt{\frac{H}{R}}}. \]

The mass lands behind the tower in the direction opposite to rotation. The consistency condition is \[ \frac{|d_x|}{H}=\frac{2\sqrt2}{3}\sqrt{\frac{H}{R}}\ll1, \] as expected for \(H\ll R\).

B.7

For a tall tower, use an inertial frame and the full circular geometry. The constant-acceleration approximation of B.6 no longer applies.

Let \[ r=R-H \] be the distance of the release point from the station’s centre, with \(0<r<R\). Choose inertial coordinates so that release occurs at \((r,0)\), with tangential velocity \((0,\Omega r)\).

After release, no real force acts on the mass, so its trajectory is a straight line: \[ X(t)=r, \qquad Y(t)=\Omega rt. \] It first reaches the rim when \[ X^2+Y^2=R^2. \] Writing \(u=\Omega t_f\), this gives \[ r^2(1+u^2)=R^2, \qquad u=\sqrt{\frac{R^2}{r^2}-1}. \]

At impact, the angular position of the mass is \[ \alpha=\arctan u, \qquad 0<\alpha<\frac{\pi}{2}, \] while the tower’s base has rotated through the angle \(u\).

For the mass to hit the tower’s base, their angles must agree modulo \(2\pi\): \[ u-\arctan u=2\pi n. \] For \(u>0\), \(u>\arctan u\), so \(n=0\) cannot give a nonzero tower height. Therefore \(n\ge1\), implying \[ u=2\pi n+\arctan u>2\pi. \]

Consequently, \[ \frac{R^2}{r^2}-1>4\pi^2, \qquad r<\frac{R}{\sqrt{1+4\pi^2}}. \] Since \(H=R-r\), \[ \boxed{ H>R\left(1-\frac{1}{\sqrt{1+4\pi^2}}\right) \simeq0.8428R}. \] For the station identified in B.5, this is \(H>0.4214R_E\).

One can also obtain the smallest exact return height. The function \[ f(u)=u-\arctan u \] is strictly increasing for \(u>0\), because \[ f'(u)=\frac{u^2}{1+u^2}>0. \] Thus the smallest return corresponds to \(n=1\). Solving \[ u_1-\arctan u_1=2\pi \] gives \[ u_1\simeq7.72525. \] Therefore \[ \boxed{ H_{\min} =R\left(1-\frac{1}{\sqrt{1+u_1^2}}\right) \simeq0.871625R}. \] This sharper value is consistent with the requested lower bound.

B.8

Continue with \(+x\) along rotation and \(+y\) inward. The freely sliding support supplies no horizontal restoring force.

Use the local flat-floor approximation and, as instructed, neglect the \(y\)-directed Coriolis force generated by horizontal motion. The vertical equation remains \[ \ddot y+\omega^2y=0, \qquad \omega^2=\frac{k}{m}-\Omega^2. \] The initial conditions are \[ y(0)=-d,\qquad \dot y(0)=0, \qquad x(0)=0,\qquad \dot x(0)=0. \] Hence \[ \boxed{y(t)=-d\cos(\omega t)}. \]

The horizontal motion is driven by the Coriolis force associated with the vertical velocity: \[ \ddot x=2\Omega\dot y. \] Integrating once, \[ \dot x=2\Omega\bigl[y(t)-y(0)\bigr] =2\Omega d\bigl[1-\cos(\omega t)\bigr]. \] Integrating again, \[ \boxed{ x(t)=2\Omega d\left[t-\frac{\sin(\omega t)}{\omega}\right]}. \]

To describe the sketch, define \[ a=\frac{2\Omega d}{\omega}, \qquad \tau=\omega t, \qquad P=\frac{2\pi}{\omega}. \] Then \[ x=a(\tau-\sin\tau), \qquad y=-d\cos\tau. \] This is a cycloid with independently scaled horizontal and vertical coordinates.

One complete arch has the following points:

Time \(x/a\) \(y/d\) Feature
\(0\) \(0\) \(-1\) Initial cusp; mass at rest
\(P/4\) \(\pi/2-1\) \(0\) Upward crossing of equilibrium height
\(P/2\) \(\pi\) \(1\) Highest point; horizontal tangent
\(3P/4\) \(3\pi/2+1\) \(0\) Downward crossing of equilibrium height
\(P\) \(2\pi\) \(-1\) Next cusp; mass momentarily at rest

The complete sketch consists of these arches repeated towards increasing \(x\):

  • The vertical amplitude is \(d\), with \(y\) ranging from \(-d\) to \(+d\).
  • Cusps occur at \((2\pi n a,-d)\), with vertical limiting tangents.
  • Upper turning points occur at \(((2n+1)\pi a,d)\), with horizontal tangents.
  • The distance between successive cusps is \[ L=2\pi a=\frac{4\pi\Omega d}{\omega}. \]
  • Since \[ \dot x=2\Omega d(1-\cos\omega t)\ge0, \] the trajectory progresses continuously in \(+x\), without loops.
  • The mean horizontal drift velocity is \(2\Omega d\); the horizontal oscillation about that uniform drift has amplitude \(a\).

These expressions apply while the stated neglect of the vertical Coriolis force and the local flat-floor approximation remain valid.

The solution was checked against the official solution.