Subfield: Mechanics

- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: Consider a toilet paper roll with some length of it hanging off as shown. The toilet paper roll rests on a cylindrical pole of radius $r=1 \mathrm{~cm}$ and the coefficient of static friction between the role and the pole is $\mu=0.3$.

<img_4274>

The length of the paper hanging off has length $\ell=30 \mathrm{~cm}$ and the inner radius of the roll is $R_{i}=2 \mathrm{~cm}$. The toilet paper has thickness $s=0.1 \mathrm{~mm}$ and mass per unit length $\lambda=5 \mathrm{~g} / \mathrm{m}$. What is the minimum outer radius $R_{o}$ such that the toilet paper roll remains static? Answer in centimeters.

Solution:
Due to the length of toilet paper hanging off, the toilet paper will be slightly tilted, in order for torques to balance. The tilt isn't shown in the diagram, since it is meant to be found. So the tilted normal force has to be compensated by the frictional force, and just before slippage,


it will look something like:

<img_4301>

Let $\theta=\arctan \mu$. So now balancing torques about the contact point, if $m$ is the mass of the toilet paper roll,

$$
m g R_{i} \sin \theta=(\ell \lambda) g\left(R_{o}-R_{i} \sin \theta\right) .
$$

And now we need $m$. The mass per unit area is $\sigma=\lambda / s$, so $m=\pi\left(R_{o}^{2}-R_{i}^{2}\right) \lambda / s$. So substituting this in,

$$
\frac{\pi\left(R_{o}^{2}-R_{i}^{2}\right) \lambda}{s} g R_{i} \sin \theta=\ell \lambda g\left(R_{o}-R_{i} \sin \theta\right)
$$

Simplifying a bit,

$$
\pi\left(R_{o}^{2}-R_{i}^{2}\right) R_{i} \sin \theta=s \ell\left(R_{o}-R_{i} \sin \theta\right) .
$$

This is a quadratic and we can just plug in the numbers and use the quadratic formula or use a graphing calculator to finish, yielding $R_{o}=2.061 \mathrm{~cm}$.

Final answer: 2.061 cm