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: Spinning Cylinder Adithya has a solid cylinder of mass $M=10 \mathrm{~kg}$, radius $R=0.08 \mathrm{~m}$, and height $H=0.20 \mathrm{~m}$. He is running a test in a chamber on Earth over a distance of $d=200 \mathrm{~m}$ as shown below. Assume that the physical length of the chamber is much greater than $d$ (i.e. the chamber extends far to the left and right of the testing area). The chamber is filled with an ideal fluid with uniform density $\rho=700 \mathrm{~kg} / \mathrm{m}^{3}$. Adithya's cylinder is launched with linear velocity $v=10 \mathrm{~m} / \mathrm{s}$ and spins counterclockwise with angular velocity $\omega$. Adithya notices that the cylinder continues on a horizontal path until the end of the chamber. Find the angular velocity $\omega$. Do not neglect forces due to fluid pressure differences. Note that the diagram presents a side view of the chamber (i.e. gravity is oriented downwards with respect to the diagram).
<img_4281>

Assume the following about the setup and the ideal fluid:

- fluid flow is steady in the frame of the center of mass of the cylinder
- the ideal fluid is incompressible, irrotational, and has zero viscosity
- the angular velocity of the cylinder is approximately constant during its subsequent motion

Hint: For a uniform cylinder of radius $R$ rotating counterclockwise at angular velocity $\omega$ situated in an ideal fluid with flow velocity $u$ to the right far away from the cylinder, the velocity potential $\Phi$ is given by

$$
\Phi(r, \theta)=u r \cos \theta+u \frac{R^{2}}{r} \cos \theta+\frac{\Gamma \theta}{2 \pi}
$$

where $(r, \theta)$ is the polar coordinate system with origin at the center of the cylinder. $\Gamma$ is the circulation and is equal to $2 \pi R^{2} \omega$. The fluid velocity is given by

$$
\mathbf{v}=\nabla \Phi=\frac{\partial \Phi}{\partial r} \hat{\mathbf{r}}+\frac{1}{r} \frac{\partial \Phi}{\partial \theta} \hat{\theta}
$$

Solution:
We will work in the reference frame of the center of mass of the cylinder because the fluid flow is steady in this reference frame. The key intuition here is that the magnitude of the fluid velocity above the cylinder will be higher on the top because the tangential velocity of the cylinder is in the same direction as the velocity of the fluid on the top. By Bernoulli's principle, this means that the pressure on the top is lower than the pressure on the bottom, which will create a lift force on the cylinder.
With the given theory, we can model this quantitatively. In our chosen reference frame, the water



moves with velocity $v$ to the left. The velocity potential around a cylinder with radius $R$ is

$$
\Phi(r, \theta)=-v r \cos \theta-v \frac{R^{2}}{r} \cos \theta+R^{2} \omega \theta
$$

Therefore, we find

$$
\mathbf{v}=\nabla \Phi=\frac{\partial \Phi}{\partial r} \hat{\mathbf{r}}+\frac{1}{r} \frac{\partial \Phi}{\partial \theta} \hat{\theta}=-v\left(1-\frac{R^{2}}{r^{2}}\right) \cos \theta \hat{\mathbf{r}}+\left(v\left(1+\frac{R^{2}}{r^{2}}\right) \sin \theta+R \omega\right) \hat{\theta}
$$

As expected from boundary conditions, the radial velocity vanishes when $r=R$. Furthermore, on the surface of the cylinder, we have the tangential velocity of the fluid is $2 v \sin \theta+R \omega$ in the counterclockwise direction. Consider points on the cylinder at angles $\theta$ and $-\theta$. By Bernoulli's principle (ignoring the height difference which will be accounted with the buoyant force),

$$
p_{-\theta}-p_{\theta}=\frac{1}{2} \rho\left((2 v \sin \theta+R \omega)^{2}-(-2 v \sin \theta+R \omega)^{2}\right)=4 \rho v R \omega \sin \theta
$$

If we integrate this result along the surface of the cylinder, we can find the lift force per unit length. Note that only the vertical components of the pressure will matter as the horizontal components cancel due to symmetry. The vertical component of the pressure difference is then $4 \rho v r_{0} \omega \sin ^{2} \theta$. Thus, the lift force per unit length is

$$
\frac{F_{\text {lift }}}{H}=\int_{0}^{\pi} 4 \rho v R \omega \sin ^{2}(\theta)(R d \theta)=2 \pi \rho \omega v R^{2}
$$

The total left force is

$$
F_{\text {lift }}=2 \pi \rho R^{2} H \omega v \text {. }
$$

The gravitational force is $M g$, and the buoyant force is $\pi R^{2} H \rho g$. Therefore, we must have

$$
\pi R^{2} H \rho g+2 \pi \rho R^{2} H \omega v=M g
$$

Solving for $\omega$, we obtain

$$
\omega=\frac{M g}{2 \pi R^{2} H \rho v}-\frac{g}{2 v}
$$

$1.25 \mathrm{~s}^{-1}$

Final answer: $1.25$ $\mathrm{s}^{-1}$