Question

What keeps a cloud from falling? A cumulus (fair-weather) cloud consists of tiny water droplets of average radius \(5.0 \mu \mathrm{m} .\) Find the terminal velocity for these droplets at \(20^{\circ} \mathrm{C},\) assuming viscous drag. (Besides the viscous drag force, there are also upward air currents called thermals that push the droplets upward. (tutorial: rain drop)

Short answer

Answer: The terminal velocity of water droplets in a cumulus cloud at \(20^{\circ} \mathrm{C}\) is approximately \(2.69 \times 10^{-4} \frac{\text{m}}{\text{s}}\).

Step-by-step solution

1. Find the gravitational force acting on a water droplet

First, we need to find the gravitational force acting on a single water droplet. The gravitational force is given by the formula: \(F_g = m \times g\) where \(F_g\) is the gravitational force, \(m\) is the mass of the water droplet, and \(g\) is the acceleration due to gravity (approximately \(9.81 \frac{\text{m}}{\text{s}^2}\)). We must find the mass of a water droplet, which can be determined using the volume and density of water. The volume \(V\) of a sphere is given by: \(V = \frac{4}{3} \pi r^3\) where \(r\) is the radius of the droplet, and the density \(\rho\) of water is approximately \(1000 \frac{\text{kg}}{\text{m}^3}\). The mass of the droplet is then: \(m = \rho V = \rho \cdot \frac{4}{3} \pi r^3\)

2. Find the viscous drag force using Stokes' law

The viscous drag force acting on the droplet can be estimated using Stokes' law, which states: \(F_d = 6 \pi \eta r v_t\) where \(F_d\) is the viscous drag force, \(\eta\) is the dynamic viscosity of air, \(r\) is the radius of the droplet, and \(v_t\) is the terminal velocity. The dynamic viscosity of air at \(20^{\circ} \mathrm{C}\) is approximately \(1.82 \times 10^{-5} \frac{\text{kg}}{\text{m s}}\).

3. Equate the gravitational force and viscous drag force

At terminal velocity, the droplet is no longer accelerating, and the gravitational force acting on it is equal to the viscous drag force, so: \(F_g = F_d \Rightarrow m \cdot g = 6 \pi \eta r v_t\)

4. Solve for terminal velocity

Now, we need to solve the equation for terminal velocity. Substituting the mass from step 1, we get: \(\rho \cdot \frac{4}{3} \pi r^3 g = 6 \pi \eta r v_t\) Now, we can simplify the equation and solve for \(v_t\): \(v_t = \frac{\rho \cdot \frac{4}{3} \pi r^3 g}{6 \pi \eta r} = \frac{2}{9} \frac{\rho r^2 g}{\eta}\) Plug in the given values for the radius of water droplets, temperature-related dynamic viscosity, and water density: \(v_t = \frac{2}{9} \frac{(1000 \frac{\text{kg}}{\text{m}^3}) (5.0 \times 10^{-6} \text{m})^2 (9.81 \frac{\text{m}}{\text{s}^2})}{1.82 \times 10^{-5} \frac{\text{kg}}{\text{m s}}} \approx 2.69 \times 10^{-4} \frac{\text{m}}{\text{s}}\) So, the terminal velocity of water droplets in a cumulus cloud at \(20^{\circ} \mathrm{C}\) is approximately \(2.69 \times 10^{-4} \frac{\text{m}}{\text{s}}\).