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: A square based pyramid (that is symmetrical) is standing on top of a cube with side length $\ell=10 \mathrm{~cm}$ such that their square faces perfectly line up. The cube is initially standing still on flat ground and both objects have the same uniform density. The coefficient of friction between every surface is the same value of $\mu=0.3$. The cube is then given an initial speed $v$ in some direction parallel to the floor. What is the maximum possible value of $v$ such that the base of the pyramid will always remain parallel to the top of the cube? Answer in meters per second.

<img_4249>

Solution:
Let $x$ be the relative displacement of the two objects. Then:
$$
v_{i}^{2}=2 a x
$$

where the relative acceleration is:

$$
a=\frac{2 h}{3 \ell} g \mu
$$

work The acceleration of pyramid is $g \mu$ and the acceleration of cube is:

$$
\rho \ell^{3} a=\rho\left(\ell^{3}+\frac{\ell^{2} h}{3}\right) g \mu+\rho \frac{\ell^{2} h}{3} g \mu \Longrightarrow a=\frac{3 \ell+2 h}{3 \ell} g \mu
$$

Therefore, the relative acceleration is:

$$
a=\frac{6 \ell+2 h}{3 \ell} g \mu
$$

We want to direct the motion of the cube diagonally (such that horizontal sides of the cube form a 45 degree angle with the displacement). Initially, we may think that we need to let $x=\frac{\sqrt{2}}{2} \ell$ but it can start tipping before that. Moving into a non-inertial reference frame for the pyramid, we see that the effective gravity needs to point towards the back corner of the cube, so it needs to satisfy the criteria

$$
\frac{h_{\mathrm{cm}}}{\frac{\sqrt{2}}{2} \ell-x}=\frac{m g}{m g \mu} \Longrightarrow x=\frac{\sqrt{2}}{2} \ell-\mu \frac{1}{4} h
$$



Here I used the fact that the center of mass was $1 / 4$ of the way up. Substituting, we get:

$$
v=\sqrt{2\left(\frac{6 \ell+2 h}{3 \ell} g \mu\right)\left(\frac{\sqrt{2}}{2} l-\frac{\mu}{4} h\right)}=\sqrt{2 g \mu\left(2+\frac{2 h}{3 \ell}\right)\left(\frac{\sqrt{2}}{2} \ell-\frac{\mu}{4} h\right)}
$$

The maximum $v$ occurs at

$$
h / \ell=\frac{-3+\frac{2 \sqrt{2}}{\mu}}{2}
$$

giving a maximum $v$ of $v_{\max }=1.07 \mathrm{~m} / \mathrm{s}$.

Final answer: 1.07 m/s