Very small objects, such as dust particles, experience a linear drag force, D =

Question

Very small objects, such as dust particles, experience a linear drag force, D  = (bv, direction opposite the motion), where b is a constant. For a sphere of radius R, the drag constant can be shown to be b=6R, where is the viscosity of the gas.

Suppose a gust of wind has carried a 48-m-diameter dust particle to a height of 330 m . If the wind suddenly stops, how long will it take the dust particle to settle back to the ground? Dust has a density of 2700 kg/m3, the viscosity of 25C air is 2.0×105Ns/m2, and you can assume that the falling dust particle reaches terminal speed almost instantly.

Express your answer using two significant figures.

(Answer in Minutes)

Explanation / Answer

Dia of dust particle = 48 micro m at height, h = 330 m

density = 2700 kg/m^3

viscocity = 2.0 x 10^-5 N.s/m^2

Now -

Drag force on a particle, D = b*Vt

=> Vt = D / b

Again, b = 6*pi*eta*R

And, volume of particle, V = (4/3)*pi*R^3 = (4/3)*3.141*(24x10^-6)^3 = 5.79x10^-14 m^3

So, mass, m = V*density = 5.79x10^-14 * 2700 = 1.563 x 10^-10 kg.

Therefore,

Vt = mg / (6*pi*eta*R) = (1.563 x 10^-10*9.81) / (6*3.141*2x10^-5*24x10^-6) = 0.169 m/s

Therefore, time taken by the particle to drop from 330 m -

t = 330/0.169 = 1953 s = 32.5 min.

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