Subfield: Modern Physics


Question: In general, we can describe the quadratic drag on an object by the following force law:
$$
F_{D}=\frac{1}{2} C_{D} \rho A v^{2}
$$

where $A$ is the cross-sectional area of the object exposed to the airflow, $v$ is the speed of the object in a fluid, and $C_{D}$ is the drag coefficient, a dimensionless quantity that varies based on shape.

Another useful quantity to know is the Reynold's number, a dimensionless quantity that helps predict



fluid flow patterns. It is given by the formula:

$$
\operatorname{Re}=\frac{\rho v L}{\mu}
$$

where $\rho$ is the density of the surrounding fluid, $\mu$ is the dynamic viscosity of the fluid, and $L$ is a reference length parameter that varies based on each object. For a smooth ${ }^{1}$ sphere traveling in a fluid, its diameter serves as the reference length parameter.

![](https://cdn.mathpix.com/cropped/2023_12_21_a44161aecfd4a6e63235g-1.jpg?height=574&width=894&top_left_y=556&top_left_x=599)

A logarithmic graph of $C_{D}$ vs Re of a sphere from the NASA Glenn Research Center.

The relationship between the drag coefficient and the Reynold's number holds significant importance. Due to the complexity of fluid dynamics, empirical data is commonly used, as depicted in the figure provided above. Notably, the figure indicates a significant decrease in the drag coefficient around $\operatorname{Re} \approx 4 \times 10^{5}$. This phenomenon, known as the drag crisis, occurs when a sphere transitions from laminar to turbulent flow, resulting in a broad wake and high drag.

Let's consider a smooth ball with a radius of $0.2 \mathrm{~m}$ and a mass of $0.1 \mathrm{~kg}$ dropped in air with a constant density of $\rho=1.255 \mathrm{~kg} / \mathrm{m}^{3}$. It is found that at velocity $5 \mathrm{~m} / \mathrm{s}$, the Reynold's number of the ball is $3.41 \cdot 10^{5}$. If the ball is dropped from rest, it approaches a stable terminal velocity $v_{1}$. If the ball is thrown downwards with enough velocity, it will experience turbulence, and approach a stable terminal velocity $v_{2}$. Find $\Delta v=v_{2}-v_{1}$. Ignore any terminal velocities found for Reynold numbers less than an order of magnitude $10^{-1}$.

Note: This problem is highly idealized as it assumes the atmosphere has air of constant density and temperature. In reality, this is not true!

Solution:
Terminal velocity exists when the net force is 0 . Using $v=\frac{\mu \cdot \operatorname{Re}}{2 \rho r}$ where $L=2 r$, we find that
$$
\frac{1}{2} \rho_{a} C_{D}\left(\pi r^{2}\right)\left(\frac{\mu \cdot \mathrm{Re}}{2 \rho r}\right)^{2}=m g-\rho_{a} g\left(\frac{4}{3} \pi r^{3}\right)
$$


Since $\rho=\frac{m}{4 \pi r^{3} / 3}=2.98 \mathrm{~kg} / \mathrm{m}^{3}$ is on the same order as $\rho_{a}=1.255 \mathrm{~kg} / \mathrm{m}^{3}$, the buoyant force must be accounted for and is non-negligible. We can rearrange to find that
$$
C_{D} \operatorname{Re}^{2}=\frac{8 \rho_{a}}{\pi \mu^{2}}\left(m g-\frac{4}{3} \rho_{a} g \pi r^{3}\right)
$$
Using $x$ as $C_{D}$ and $y$ as Re, we can plot an equation $x y^{2}=$ const on the $C_{D}$ vs Re graph. There, we can find three intersections.

![](https://cdn.mathpix.com/cropped/2023_12_21_aa674442fff889231bc7g-1.jpg?height=721&width=827&top_left_y=596&top_left_x=649)

The intersection in the middle is not stable. So we find the intersections of the other two to be $\operatorname{Re}_{1} \approx 2.6 \times 10^{5}$ and $\operatorname{Re}_{2}=6 \times 10^{5}$. Hence, $v_{1}=3.81 \mathrm{~m} / \mathrm{s}$ and $v_{2}=8.79 \mathrm{~m} / \mathrm{s}$, meaning $\Delta v=4.98 \mathrm{~m} / \mathrm{s}$.

Final answer: $4.98$ m/s