Thought Of The Day_10_Drops From Clouds

#FluidMechanics #Mechanics #Aerodynamics #Viscosity

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Let us derive a simplistic formula for terminal velocity:

Assume a spherical liquid drop of radius $r$ , density $\rho$ and coefficient of viscosity $\eta$ . It is falling through air of density $\sigma$ .

The forces acting on it are:

Downward - Gravity
Upward - Viscous drag (in accordance with Stokes' Law)
Upward - Buoyant force (weight of displaced air)

At terminal velocity, our drop is in equilibrium:

$\dfrac{4}{3} \pi r^3 \rho g = 6 \pi \eta r V_T + \dfrac{4}{3} \pi r^3 \sigma g$

Simplifying this, we obtain the expression for terminal velocity:

$\boxed{V_T = \dfrac{2 r^2 g}{9 \eta} (\rho -\sigma)}$

This shows that a drop with larger radius will reach the ground at a higher terminal velocity.

Raj Magesh - 6 years ago

Correct the key here is that they will attain the terminal velocity..!!
We can have a bonus question now..
If two identical drops fall from clouds at different heights, which will hit the groud with greater speed, the one from the higher cloud or the other one?

Rohit Gupta - 6 years ago

Assuming they both manage to reach terminal velocity (a very reasonable assumption), they will both hit the ground at the same speed (terminal velocity).

Raj Magesh - 6 years ago

@Raj Magesh – Correct!! Bravo..!!!

Rohit Gupta - 6 years ago

Assuming drops to be spheres, just use that odd equation for terminal velocity of bodies.

Raghav Vaidyanathan - 6 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$