Question

Calculate the speed (in m/s) a spherical rain drop would achieve
falling from 2.10 km in the absence of air drag and with air drag.
Take the size across of the drop to be 9 mm, the density to be 1.00
✕ 10^{3} kg/m^{3}, and the surface area to be
*π**r*^{2}. (Assume the density of air is
1.21 kg/m^{3}.)

(a)

in the absence of air drag

202.985 m/s (correct)

(b)

with air drag

9.8687 m/s (incorrect)

Answer #1

(a) Use energy conservation -

1/2 m v^2 = m g h

v = (2 g h)^1/2 = (2 * 9.81 * 2100)^1/2 = 202.98 m/s

(b) Use dynamics -

m x" = m g - k x'^2

limit speed is

v = (m g / k)^1/2

where

k = 1/2 ρ A C

ρ = air density

A = area [π r^2] = [3.14x(4.5x10^-3)^2] = 63.59 x 10^-6 m^2

C = coefficient of hydraulic resistance [0.47 for a sphere]

m = ρ' V = 10^3 x (4/3)*3.14*(4.5x10^-3)^3 = 381.51 x 10^-6 kg [ρ'
= 10^3 kg/m^3 (drop density)]

k = (1/2)x1.21x63.59x10^-6x0.47 = 1.808 x 10^-5 kg/m

So, v = (381.51 x 10^-6 x 9.81 / 1.808 x 10^-5)^1/2 = 14.4 m/s

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