Question

Very small objects, such as dust particles, experience a linear drag force, D⃗ 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=6πηR, where η is the viscosity of the gas.

Suppose a gust of wind has carried a 52-μm-diameter dust particle to a height of 260 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/m^3, the viscosity of 25.C air is 2.0×10−5N⋅s/m^2, and you can assume that the falling dust particle reaches terminal speed almost instantly. Answer in minutes

Answer 1

at terminal speed, drag force is equal to weight.

so,

6nRv = mg

so,

v = mg / 6nR

Now, we need to find mass of particle

mass = density * volume ( treat the particle as sphere)

mass = 2700 * 4/3 * * 26e-6³

mass = 1.9877e-10 Kg

so,

v = 1.9877e-10 * 9.8 / 6 * 2e-5 * 26e-6

v = 0.198744 m/s

so,

t = d / v

t = 260 / 0.198744

t = 1308.2 seconds

or

t = 21.8 minutes