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# Thought Of The Day_10_Drops From Clouds

Note by Rohit Gupta
2 years, 7 months ago

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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:

1. Downward - Gravity
2. Upward - Viscous drag (in accordance with Stokes' Law)
3. 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.

- 2 years, 7 months 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?

- 2 years, 7 months 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).

- 2 years, 7 months ago

Correct!! Bravo..!!!

- 2 years, 7 months ago

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

- 2 years, 7 months ago

Comment deleted Jun 09, 2015

If drag force is more then weight is more as well... Which one is gonna dominate and why??

- 2 years, 7 months ago