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 Somewhere inside the wooden cylinder, the wood has been replaced by a metal disk of radius $r_{2}$ and thickness $h_{2}$.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

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 $d_{x} \ll H$.

Solution:
- Velocity $v_{x}$

Equation for Coriolis force with correct velocity:

$$
F_{C}(t)=2 m \omega_{s s}^{2} R t \omega_{s s}=2 m \omega_{s s}^{3} R t
$$

Integrate this, or realize that it is like uniform acceleration for the velocity:

$$
v_{x}(t)=\omega_{s s}^{3} R t^{2}
$$

plug in correct value for

$$
t=\sqrt{2 H / \omega_{s s}^{2} R}
$$

overall correct result

$$
v_{x}=2 H \omega_{s s}
$$

- The displacement $d_{x}$ :

Integrate $v_{x}(t)$ :

$$
d_{x}=\frac{1}{3} R \omega_{s s}^{3} t^{3}
$$

Plug in value for $t$

$$
d_{x}=\frac{1}{3} R \omega_{s s}^{3}\left(2 H / \omega_{s s}^{2} R\right)^{3 / 2}=\frac{1}{3} 2^{3 / 2} H^{3 / 2} R^{-1 / 2}=\frac{1}{3} \sqrt{\frac{8 H^{3}}{R}}
$$
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 $v_{x}=\omega_{s s}(R-H)$.

When the mass lands, the $x$-direction has been rotated by $\phi$ so the new horizontal velocity component is then

$$
\omega_{s s}(R-H) \cos \phi
$$

(Student may also write $\cos \alpha$ instead of $\cos \phi$, since $d_{x} \ll H$.)

$$
\cos \phi=\frac{R-H}{R}=1-\frac{H}{R}
$$

Transforming to the rotating reference frame, one needs to subtract $\omega_{s s} R$.

Finally in the reference frame of the astronauts

$$
v_{x}=\omega_{s s} R\left(1-\frac{H}{R}\right)^{2}-\omega_{s s} R \approx \omega_{s s} R\left(1-2 \frac{H}{R}\right)-\omega_{s s} R=-2 \omega_{s s} H
$$

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 $v_{x}$

$$
\begin{gathered}
d_{x}=(\alpha-\phi) R \\
\phi=\arccos \left(1-\frac{H}{R}\right) \\
\alpha=\omega_{s s} t
\end{gathered}
$$

where $t$ is the fall time of the mass, which is given by

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

(see solution to $\mathrm{B} 7$ )

Writing $\xi \equiv H / R$ this means

$$
d_{x}=\left[\frac{\sqrt{1-(1-\xi)^{2}}}{1-\xi}-\arccos (1-\xi)\right] R
$$

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:

$$
d_{x}=\frac{2}{3} \sqrt{\frac{2 H^{3}}{R}}
$$

Final answer: $2 H \omega_{s s}$,$\frac{1}{3} \sqrt{\frac{8 H^{3}}{R}}$
$-2 H \omega_{s s}$,$\frac{1}{3} \sqrt{\frac{8 H^{3}}{R}}$