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Условие Part A. The Hidden Disk (3.5 points)Part B. Rotating Space Station (6.5 points)Отговори Решение

IPhO 2016, Теория — Задача 1. Two Problems in Mechanics

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IPhO 2016 — Theory Q1 — Two Problems in Mechanics (10 points) · 10 т.

Условие

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 Somewhere inside the wooden cylinder, the wood has been replaced by a metal disk of radius $r_2$ and thickness $h_2$ .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\ . \qquad (1)

The goal is to determine $r_2, h_2$ and $d$ , through indirect measurements.

a) Side view: a large light-grey disk of radius r1 with centre S and a smaller dark-grey disk of radius r2 with centre B at distance d from S. b) View from above: rectangle of width h1 with dashed axes S and B, radius r1 from the left edge to S, the embedded metal disk of thickness h2 and radius r2 around axis 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 $\Theta$ (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.

The cylinder with the hidden disk resting on a base inclined by the angle Θ; the vertical through the axis S and the line from S to the centre of the metal disk make the angle ϕ. — Figure 2: Cylinder on an inclined base.

A.1 Find an expression for $b$ as a function of the quantities (1), the angle $\phi$ and the tilting angle $\Theta$ of the base. [0,8 т.]

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

The cylinder suspended at its symmetry axis S from a rigid rod hanging from a horizontal support; the line to the metal disk is turned by the small angle φ from the vertical. — 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 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. [0,5 т.]

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). You may also include $r_2$ and $h_2$ as variables in your expression, as they will be calculated in subtask A.5. [0,4 т.]

A.4 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. [0,7 т.]

A.5 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 $r_2$ . [1,1 т.]

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 $\omega_{ss}$ does the space station rotate so that the astronauts experience the same gravity $g_E$ as on the Earth's surface? [0,5 т.]

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 $g_E$ , what would be the angular oscillation frequency $\omega_E$ that a person on Earth would measure? [0,2 т.]

B.3 What angular oscillation frequency $\omega$ does Alice measure on the space station? [0,6 т.]

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.

A circle (the space station wheel) with its centre marked, radius R drawn to the rim, a curved arrow labelled ω_ss showing the rotation, and two stick figures standing on the inner side of the rim at the bottom. — Figure 4: Space station

B.4 Derive an expression of the gravity $g_E(h)$ for small heights $h$ above the surface of the Earth and compute the oscillation frequency $\tilde{\omega}_E$ of the oscillating mass (linear approximation is enough). Denote the radius of the Earth by $R_E$ . Neglect the rotation of Earth. [0,8 т.]

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 $\omega$ match the oscillation frequency $\tilde{\omega}_E$ on the Earth? Express your answer in terms of $R_E$ . [0,3 т.]

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}\ . \qquad (2)

In terms of the scalar quantities you may use

F_C = 2mv\omega_{ss}\sin\phi\ , \qquad (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 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 $d_x \ll H$ . [1,1 т.]

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 Find a lower bound for the height of the tower for which it can happen that $d_x = 0$ . [1,3 т.]

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.

A ring on a horizontal rod with a vertical spring hanging from it; the dashed level y = 0 marks the equilibrium point and the mass is pulled a distance d below it. — Figure 5: Setup.

B.8 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. [1,7 т.]

Alice and Bob continue to argue.

Отговори

Решение

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

Part A. The Hidden Disk (3.5 points)

Solution A1: [0.8]

Geometric solution: use that torque with respect to point of contact is 0 $\Rightarrow$ center of gravity has to be vertically above point of contact.

\sin\phi = \frac{D}{b} \qquad \mathbf{0.3}

\sin\Theta = \frac{D}{r_1} \qquad \mathbf{0.3}

Here $D$ may be called another name. Solve this:

\sin\phi = \frac{r_1}{b}\sin\Theta\ \Rightarrow\ b = \frac{r_1\sin\Theta}{\sin\phi} \qquad \mathbf{0.2}

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

Solution A2: [0.5]

Write some equation of the form $\ddot{\varphi} = -\omega^2\varphi$ 0.1

Writing an equation of the form $\varphi = A\cos\omega t$ is also correct.
Two solutions:

Kinetic energy: $\frac{1}{2}I_S\dot{\varphi}^2$ and potential energy: $-bMg\cos\varphi$ . Total energy is conserved, and differentiation w.r.t. time gives the equation of motion.
Angular equation of motion from torque, $\tau = I_S\ddot{\varphi} = -Mgb\sin\varphi$ .

Correct equation (either energy conservation or torque equation of motion) 0.3
Final answer

T = 2\pi\sqrt{\frac{I_S}{Mgb}}\ \Rightarrow\ I_S = \frac{MgbT^2}{4\pi^2} \qquad \mathbf{0.1}

(Derivation:

\Rightarrow \ddot{\varphi} = -\frac{bMg}{I_S}\sin\varphi \simeq -\frac{bgM}{I_S}\varphi

so that

\omega^2 = \frac{bgM}{I_S}

)

Solution A3: [0.4]

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

b = \frac{dM_2}{M_1 + M_2} \qquad \mathbf{0.2}

correct solution:

d = \frac{bM}{\pi h_2 r_2^2(\rho_2 - \rho_1)} \qquad \mathbf{0.2}

Solution A4: [0.7]

correct answer for moment of inertia of homogeneous disk

I_1 = \frac{1}{2}\pi h_1\rho_1 r_1^4 \qquad \mathbf{0.2}

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:

I_2 = \frac{1}{2}\pi h_2(\rho_2 - \rho_1)r_2^4 \qquad \mathbf{0.2}

Using Steiner’s theorem:

I_S = I_1 + I_2 + d^2\pi r_2^2 h_2(\rho_2 - \rho_1) \qquad \mathbf{0.1}

correct solution:

I_S = \frac{1}{2}\pi h_1\rho_1 r_1^4 + \frac{1}{2}\pi h_2(\rho_2 - \rho_1)r_2^4 + \frac{b^2M^2}{\pi r_2^2 h_2(\rho_2 - \rho_1)} \qquad \mathbf{0.2}

In terms of $d$ rather than $b$ gives 0.1pts rather than 0.2pts for the final answer: 0.1

I_S = \frac{1}{2}\pi h_1\rho_1 r_1^4 + \frac{1}{2}\pi h_2(\rho_2 - \rho_1)r_2^4 + d^2\pi r_2^2 h_2(\rho_2 - \rho_1)

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 = \pi r_1^2 h_1\rho_1 + \pi r_2^2 h_2(\rho_2 - \rho_1)\ . \qquad \mathbf{0.3}

solve $I_S$ for $r_2^2$ :

r_2^2 = \frac{2}{M - \pi r_1^2 h_1\rho_1}\left(I_S - \frac{1}{2}\pi h_1\rho_1 r_1^4 - b^2\frac{M^2}{M - \pi r_1^2 h_1\rho_1}\right) \qquad \mathbf{0.4}

replace $I_S$ by $T$ :

I_S = \frac{MgbT^2}{4\pi^2} \qquad \mathbf{0.1}

solve correctly for $r_2$ :

r_2 = \sqrt{\frac{2}{M - \pi r_1^2 h_1\rho_1}\left(M\frac{bgT^2}{4\pi^2} - \frac{1}{2}\pi h_1\rho_1 r_1^4 - b^2\frac{M^2}{M - \pi r_1^2 h_1\rho_1}\right)} \qquad \mathbf{0.1}

write down an equation for $h_2$ along the lines of $M = \pi r_1^2\rho_1 h_1 + \pi r_2^2(\rho_2 - \rho_1)h_2$ and solve it correctly:

h_2 = \frac{M - \pi r_1^2\rho_1 h_1}{\pi r_2^2(\rho_2 - \rho_1)} \qquad \mathbf{0.2}

Part B. Rotating Space Station (6.5 points)

Solution B1: [0.5]

An equation for the centrifugal force along the lines of

F_{ce} = m\omega^2 r \qquad \mathbf{0.1}

Balancing the forces, correct equation

g_E = \omega_{ss}^2 R \qquad \mathbf{0.2}

Correct solution

\omega_{ss} = \sqrt{g_E/R} \qquad \mathbf{0.2}

Solution B2: [0.2]

Realize that result is independent of $g_E$ 0.1

Correct result:

\omega_E = \sqrt{k/m} \qquad \mathbf{0.1}

Solution B3: [0.6]

some version of the correct equation for force

F = -kx \pm m\omega_{ss}^2 x \qquad \mathbf{0.2}

getting the sign right

F = -kx + m\omega_{ss}^2 x \qquad \mathbf{0.2}

Find correct differential equation

m\ddot{x} + (k - m\omega_{ss}^2)x = 0 \qquad \mathbf{0.1}

Derive correct result

\omega = \sqrt{k/m - \omega_{ss}^2} \qquad \mathbf{0.1}

Using $g_E/R$ instead of $\omega_{ss}^2$ is also correct.

Solution B4: [0.8]

g_E(h) = -GM/(R_E + h)^2 \qquad \mathbf{0.1}

linear approximation of gravity:

g_E(h) = -\frac{GM}{R_E^2} + 2h\frac{GM}{R_E^3} + \ldots \qquad \mathbf{0.2}

Realize that $g_E = GM/R_E^2$ :

g_E(h) = -g_E + 2hg_E/R_E + \ldots \qquad \mathbf{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 + 2xmg_E/R_E \qquad \mathbf{0.2}

Find correct differential equation

m\ddot{x} + (k - 2mg_E/R_E)x = 0 \qquad \mathbf{0.1}

correct result

\tilde{\omega}_E = \sqrt{k/m - 2g_E/R_E} \qquad \mathbf{0.1}

No points are deducted if student answers with $\tilde{\omega}_E/(2\pi)$ because ”oscillation frequency” might also be interpreted as inverse period.

Solution B5: [0.3]

Write down equation

\omega_{ss}^2 = 2g_E/R_E \qquad \mathbf{0.1}

Solve

R = R_E/2 \qquad \mathbf{0.2}

If $GM/R_E^2$ rather than $g_E$ is used, give only 0.1pt.

Solution B6: [1.1]

There are several possible solutions.
Solution one – Using Coriolis force

Velocity $v_x$

Equation for Coriolis force with correct velocity:
$F_C(t) = 2m\omega_{ss}^2 Rt\omega_{ss} = 2m\omega_{ss}^3 Rt \qquad \mathbf{0.1}$
Integrate this, or realize that it is like uniform acceleration for the velocity:
$v_x(t) = \omega_{ss}^3 Rt^2 \qquad \mathbf{0.2}$
plug in correct value for
$t = \sqrt{2H/\omega_{ss}^2 R} \qquad \mathbf{0.2}$
overall correct result
$v_x = 2H\omega_{ss} \qquad \mathbf{0.1}$
The displacement $d_x$ :

Integrate $v_x(t)$ :
$d_x = \frac{1}{3}R\omega_{ss}^3 t^3 \qquad \mathbf{0.3}$
Instead of integrating, students may simply ‘average’ by taking $\frac{1}{2}$ of the final velocity. This gives a factor of $\frac{1}{2}$ instead of $\frac{1}{3}$ . Deduct a total of 0.1 pts for this. -0.1

Plug in value for $t$
$d_x = \frac{1}{3}R\omega_{ss}^3(2H/\omega_{ss}^2 R)^{3/2} = \frac{1}{3}2^{3/2}H^{3/2}R^{-1/2} = \frac{1}{3}\sqrt{\frac{8H^3}{R}} \qquad \mathbf{0.2}$

Solution two – Using inertial frame This solution is similar to the way to solve B7, but needs more complicated approximations than Solution one.

$v_x$

Here $\phi$ denotes the angle swept by the mass and $\alpha$ 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 $v_x = \omega_{ss}(R - H)$ . 0.1

When the mass lands, the $x$ -direction has been rotated by $\phi$ so the new horizontal velocity component is then
$\omega_{ss}(R - H)\cos\phi \qquad \mathbf{0.1}$
(Student may also write $\cos\alpha$ instead of $\cos\phi$ , since $d_x \ll H$ .)
$\cos\phi = \frac{R - H}{R} = 1 - \frac{H}{R} \qquad \mathbf{0.1}$
Transforming to the rotating reference frame, one needs to subtract $\omega_{ss}R$ . 0.1

Finally in the reference frame of the astronauts
$v_x = \omega_{ss}R\left(1 - \frac{H}{R}\right)^2 - \omega_{ss}R \approx \omega_{ss}R\left(1 - 2\frac{H}{R}\right) - \omega_{ss}R = -2\omega_{ss}H \qquad \mathbf{0.2}$
The sign of the velocity depend on the choice of reference direction, so a positive sign is also correct.
$d_x$

With the notation from the calculation of $v_x$
$d_x = (\alpha - \phi)R \qquad \mathbf{0.1}$ $\phi = \arccos\left(1 - \frac{H}{R}\right)$ $\alpha = \omega_{ss}t$
where $t$ is the fall time of the mass, which is given by
$t = \frac{\sqrt{R^2 - (R - H)^2}}{\omega_{ss}(R - H)} \qquad \mathbf{0.1}$
(see solution to B7)

Writing $\xi \equiv H/R$ this means
$d_x = \left[\frac{\sqrt{1 - (1 - \xi)^2}}{1 - \xi} - \arccos(1 - \xi)\right]R \qquad \mathbf{0.3}$
which is a valid end answer to the problem. It is possible, but not necessary, to approximate this for small $\xi$ :
$\arccos(1 - \xi) \approx \sqrt{2\xi}\left(1 + \frac{\xi}{12}\right)$
which after insertion into the equation for $d_x$ and approximation of small $\xi$ yields the same result as in Solution one:
$d_x = \frac{2}{3}\sqrt{\frac{2H^3}{R}}$
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 $d_x$
The mass travels the distance $l$ , and during the fall the space station rotates by $\phi$ , see Figure 2. According to the intersecting chord theorem,

l^2 = H(2R - H) \qquad \mathbf{0.1}

The rotated angle is $\phi = \omega_{ss}t$ where

t = \frac{l}{R - H} \qquad \mathbf{0.1}

is the fall time. Thus

\phi = \frac{\sqrt{H(2R - H)}}{R - H} \qquad \mathbf{0.1}

\frac{d}{R} = \phi - \arcsin\frac{l}{R} = \frac{\sqrt{H(2R - H)}}{R - H} - \arcsin\sqrt{x(2 - x)} \qquad \mathbf{0.1}

Hand-drawn sketch: the rim of the station with the tower; angles ϕ and α at the centre, the tangent at the landing point, the initial velocity arrow ω_ss(R − H), the angle ϕ and the displacement d_x marked in red along the rim. — Figure 1: Notation for solution two

Figure 1: Notation for solution two

Hand-drawn circle with a vertical diameter split into 2R − H and H, a horizontal chord with halves l and l, and the angle ϕ marked. — Figure 2: Notation for solution three.

Figure 2: Notation for solution three.

Denote $x \equiv H/R$ and $y \equiv \sqrt{x(2 - x)}$ . Since

\arcsin y \approx y + \frac{y^3}{6}

one gets

\frac{d}{R} \approx y(1 + x) - y - y^3/6 = y(x - y^2/6) \approx 2xy/3 \approx 2x\sqrt{2x}/3 = \frac{2}{3}\sqrt{\frac{2H^3}{R}}

Final answer 0.1

Solution B7: [1.3]

The key is to use a non-rotating frame of reference. If the mass is released close enough to the center, its linear velocity will be small enough for the space station to rotate more than $2\pi$ before it hits the ground.
The velocity is given by

v = \omega_{ss}(R - H) \qquad \mathbf{0.1}

distance $d$ that the mass flies before hitting the space station

d^2 = R^2 - (R - H)^2 \qquad \mathbf{0.1}

use non-rotating frame of reference to obtain time $t$ until impact

t = d/v = \frac{\sqrt{R^2 - (R - H)^2}}{\omega_{ss}(R - H)} \qquad \mathbf{0.1}

Now there are several possible ways to relate $H$ and the rotated angle $\phi$ of the space station:
Solution one

t = \frac{R\sin\phi}{\omega_{ss}R\cos\phi} \qquad \mathbf{0.2}

This time must match $t = \phi/\omega_{ss}$ . Obtain the equation

\phi = \tan\phi \qquad \mathbf{0.2}

Realizing that there is an infinite number of solutions. 0.2
This equation has one trivial solution $\phi = 0$ , next solution is slightly less than $3\pi/2$ 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

\phi \approx 5\pi/2

The equation $\phi = \tan\phi$ can be solved graphically or numerically to obtain a close value ( $\phi = 7.725\ \mathrm{rad}$ ) which means

H/R = (1 - \cos\phi) \approx 0.871

Give points if the method is correct, depending on the value of $H/R$ found, according to these intervals: 0.4
$0.85 \le H/R \le 0.88$ : 0.4 pts
$0.5 \le H/R < 0.85$ : 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 $\phi$

\frac{R - H}{R} = \cos\phi \qquad \mathbf{0.2}

obtain equation of the form

\frac{H}{R} = 1 - \cos\left(\frac{\sqrt{1 - (1 - H/R)^2}}{1 - H/R}\right) \qquad \mathbf{0.2}

Figure 3 gives a plot of $f(x) = 1 - \cos\left(\frac{\sqrt{1 - (1 - x)^2}}{1 - x}\right)$ . The goal is to find an approximate solution for the second intersection. The first intersection is discarded – it is introduced because of $\cos\phi = \cos(-\phi)$ and corresponds to a situation with $H > R$ .
Realizing that there is an infinite number of solutions. 0.2

Plot on 0 ≤ x ≤ 1, 0 ≤ y ≤ 2 of f(x) = 1 − cos(√(1 − (1 − x)²)/(1 − x)), which rises to 2 near x ≈ 0.7 and then oscillates ever faster towards x = 1, together with the straight line y = x. — Figure 3: Plot of $f(H/R)$ and $H/R$

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

Plot on 0 ≤ x ≤ 1, −1 ≤ y ≤ 1 of g(x) = cos(√(1 − x²)/x), oscillating rapidly near x = 0, with a minimum −1 near x ≈ 0.3 and rising to 1 at x = 1, together with the straight line y = x. — Figure 4: Plot of $g(x)$ and $x$

Figure 4: Plot of $g(x)$ and $x$

introduce new variable $x := 1 - H/R$ , so that the equation becomes
$x = \cos(\sqrt{1 - x^2}/x) =: g(x)$
$g(x)$ 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:
$\frac{\sqrt{1 - x^2}}{x} = 5\pi/2$
give lower bound
$x = 1/\sqrt{25\pi^2/4 + 1} \Rightarrow H = R(1 - 1/\sqrt{25\pi^2/4 + 1}) \approx 0.874$
Note: the actual result is $H/R = 0.871\ldots$ .

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.

It is also possible to use $\cos\left(\frac{\sqrt{1 - x^2}}{x}\right) = \sin(1/x)$ .

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 $\omega$ , they don’t need to write $\sqrt{k/m - \omega_{ss}^2}$ explicitly. Deduct 0.1pt however if they use $k/m$ instead of $\omega$ .. -0.1

Realize that $y(t)$ is standard harmonic oscillation:

y(t) = A\cos\omega t + B \qquad \mathbf{0.1}

Give correct constants from initial conditions

y(t) = -d\cos\omega t \qquad \mathbf{0.2}

Correct expression for $v_y(t)$ :

v_y(t) = -d\omega\sin\omega t \qquad \mathbf{0.1}

Coriolis force in $x$ -direction

F_x(t) = 2m\omega_{ss}v_y(t) = -2m\omega_{ss}d\omega\sin\omega t \qquad \mathbf{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:

A = \frac{2\omega_{ss}d}{\omega} \qquad \mathbf{0.1}

Correct answer with correct initial conditions:

x(t) = \frac{2\omega_{ss}d}{\omega}\sin\omega t - 2\omega_{ss}dt \qquad \mathbf{0.2}

Sketch:

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 $y = \pm d$ 0.1

C): cusps are at distance $\Delta x = \frac{4\pi\omega_{ss}d}{\omega}$ from each other 0.2

Оригинал в Архива: 1.pdf · официални решения: 1_sol.pdf