Subfield: Mechanics

Two Problems in Mechanics

Please read the general instructions in the separate envelope before you start this problem.

Part A.The Hidden Disk

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 Somewhere inside the wooden cylinder, the wood has been replaced by a metal disk of radius $r_{2}$ and thickness $h_{2}$.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.

а)

<img_4399>

b)

<img_4400>

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.


<img_4560>

Figure 2: Cylinder on an inclined base.
Context question:
A.1 Find an expression for $b$ as a function of the quantities (1), the angle $\phi$ and the $0.8 p t$ tilting angle $\Theta$ of the base.
Context answer:
\boxed{$b=\frac{r_{1} \sin \Theta}{\sin \phi}$}


Extra Supplementary Reading Materials:

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

<img_4484>

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 $T$.
Context question:
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.
Context answer:
Kinetic energy: $\frac{1}{2} I_{S} \dot{\varphi}^{2}$ and potential energy: $-b M g \cos \varphi$. Total energy is conserved, and differentiation w.r.t. time gives the equation of motion.

Angular equation of motion from torque, $\tau=I_{S} \ddot{\varphi}=-M g b \sin \varphi$.


Correct equation (either energy conservation or torque equation of motion)

$$
T=2 \pi \sqrt{\frac{I_{S}}{M g b}} \Rightarrow I_{S}=\frac{M g b T^{2}}{4 \pi^{2}}
$$



(Derivation:

$$
\Rightarrow \ddot{\varphi}=-\frac{b M g}{I_{S}} \sin \varphi \simeq-\frac{b g M}{I_{S}} \varphi
$$

so that

$$
\omega^{2}=\frac{b g M}{I_{S}}
$$


Extra Supplementary Reading Materials:

From the measurements in questions $\mathbf{A} .1$ and $\mathbf{A . 2}$, we now want to determine the geometry and the position of the metal disk inside the wooden cylinder.
Context question:
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.
Context answer:
\boxed{$d=\frac{b M}{\pi h_{2} r_{2}^{2}(\rho_{2}-\rho_{1})}$}


Context question:
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.
Context answer:
\boxed{$I_{S}=\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}+\frac{1}{2} \pi h_{2}(\rho_{2}-\rho_{1}) r_{2}^{4}+d^{2} \pi r_{2}^{2} h_{2}(\rho_{2}-\rho_{1})$}


Context question:
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 $r_{2}$.
Context answer:
\boxed{$r_{2}=\sqrt{\frac{2}{M-\pi r_{1}^{2} h_{1} \rho_{1}}(M \frac{b g T^{2}}{4 \pi^{2}}-\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}-b^{2} \frac{M^{2}}{M-\pi r_{1}^{2} h_{1} \rho_{1}})}$ , $h_{2}=\frac{M-\pi r_{1}^{2} \rho_{1} h_{1}}{\pi r_{2}^{2}(\rho_{2}-\rho_{1})}$}


Extra Supplementary Reading Materials:

Part B.Rotating Space Station

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.
Context question:
B.1 At what angular frequency $\omega_{s s}$ does the space station rotate so that the astronauts experience the same gravity $g_{E}$ as on the Earth's surface?
Context answer:
\boxed{$\omega_{s s}=\sqrt{\frac{g_{E}}{R}}$}


Extra Supplementary Reading Materials:

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.
Context question:
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?
Context answer:
\boxed{$\omega_{E}=\sqrt{k / m}$}


Context question:
B.3 What angular oscillation frequency $\omega$ does Alice measure on the space station?
Context answer:
\boxed{$\sqrt{k / m-\omega_{s s}^{2}}$}


Extra Supplementary Reading Materials:

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.


<img_4437>

Figure 4: Space station
Context question:
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 $\tilde{\omega}_{F}$ of the oscillating mass (linear approximation is enough). Denote the radius of the Earth by $R_{E}$. Neglect the rotation of Earth.
Context answer:
\boxed{$g_{E}(h)=\frac{-G M}{(R_{E}+h)^{2}}$ , $\tilde{\omega}_{E}=\sqrt{k / m-2 g_{E} / R_{E}}$}


Extra Supplementary Reading Materials:

Indeed, for this space station, Alice does find that the spring pendulum oscillates with the frequency that Bob predicted.
Context question:
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 $R_{E}$.
Context answer:
\boxed{$R=R_{E} / 2$}


Extra Supplementary Reading Materials:

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 $\vec{\omega}_{s s}$ is given by

$$
\vec{F}_{C}=2 m \vec{v} \times \vec{\omega}_{s s} .
\tag{2}
$$

In terms of the scalar quantities you may use

$$
F_{C}=2 m v \omega_{s s} \sin \phi,
\tag{3}
$$

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.
Context question:
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 $d_{x} \ll H$.
Context answer:
\boxed{$2 H \omega_{s s}$,$\frac{1}{3} \sqrt{\frac{8 H^{3}}{R}}$}


Extra Supplementary Reading Materials:

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 $d_{x}=0$.

Question: B.7 Find a lower bound for the height of the tower for which it can happen that $d_{x}=0$.

Solution:
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

$$
v=\omega_{s s}(R-H)
$$

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

$$
d^{2}=R^{2}-(R-H)^{2}
$$

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

$$
t=d / v=\frac{\sqrt{R^{2}-(R-H)^{2}}}{\omega_{s s}(R-H)}
$$


$$
t=\frac{R \sin \phi}{\omega_{s s} R \cos \phi}
$$

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

$$
\phi=\tan \phi
$$

Realizing that there is an infinite number of solutions.

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

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

$$
v=\omega_{s s}(R-H)
$$

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

$$
d^{2}=R^{2}-(R-H)^{2}
$$

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

$$
t=d / v=\frac{\sqrt{R^{2}-(R-H)^{2}}}{\omega_{s s}(R-H)}
$$


relation between $H$ and rotated angle $\phi$

$$
\frac{R-H}{R}=\cos \phi
$$

obtain equation of the form

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

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.



<img_4452>

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

<img_4411>

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

- introduce new variable $x:=1-H / R$, so that the equation becomes

$$
x=\cos \left(\sqrt{1-x^{2}} / x\right)=: g(x)
$$

- $g(x)$ 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

$$
x=1 / \sqrt{25 \pi^{2} / 4+1} \Rightarrow H=R\left(1-1 / \sqrt{25 \pi^{2} / 4+1}\right) \approx 0.874
$$

Note: the actual result is $H / R=0.871 \ldots$

Use the same points for the numerical answer as was mentioned in solution one.

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 $\cos \left(\frac{\sqrt{1-x^{2}}}{x}\right)=\sin (1 / x)$.

Final answer: 0.871