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 Somewhere inside the wooden cylinder, the wood has been replaced by a metal disk of radius $r_{2}$ and thickness $h_{2}$.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{$I_{S}=\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}+\frac{1}{2} \pi h_{2}(\rho_{2}-\rho_{1}) r_{2}^{4}+d^{2} \pi r_{2}^{2} h_{2}(\rho_{2}-\rho_{1})$}


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 $r_{2}$.

Solution:
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:

$$
M=\pi r_{1}^{2} h_{1} \rho_{1}+\pi r_{2}^{2} h_{2}\left(\rho_{2}-\rho_{1}\right)
$$

solve $I_{S}$ for $r_{2}^{2}$ :

$$
r_{2}^{2}=\frac{2}{M-\pi r_{1}^{2} h_{1} \rho_{1}}\left(I_{S}-\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}-b^{2} \frac{M^{2}}{M-\pi r_{1}^{2} h_{1} \rho_{1}}\right)
$$

replace $I_{S}$ by $T$ :

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

solve correctly for $r_{2}$ :

$$
r_{2}=\sqrt{\frac{2}{M-\pi r_{1}^{2} h_{1} \rho_{1}}\left(M \frac{b g T^{2}}{4 \pi^{2}}-\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}-b^{2} \frac{M^{2}}{M-\pi r_{1}^{2} h_{1} \rho_{1}}\right)}
$$

write down an equation for $h_{2}$ along the lines of $M=\pi r_{1}^{2} \rho_{1} h_{1}+\pi r_{2}^{2}\left(\rho_{2}-\rho_{1}\right) h_{2}$ and solve it correctly:

$$
h_{2}=\frac{M-\pi r_{1}^{2} \rho_{1} h_{1}}{\pi r_{2}^{2}\left(\rho_{2}-\rho_{1}\right)}
$$

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