Question

Consider a very small raindrop slowly falling down in quiet air. Because of the drop’s small size and low speed, the air resistence to its fall is viscous rather than turbulent. Consequently, the resistive force is depends linearly on the drop’s velocity, R~ = −b ~v, (1) where b is a constant which depends on the size and shape of the drop and the air’s viscosity but does not depend on the drop’s weight of what it’s made of. Suppose a raindrop of weight mg = 0.36 µg × 9.8 m/s 2 reaches terminal speed vt = 0.25 m/s high above the ground. What is the numeric value of the coefficient b in eq. (1) for the air resistence to this drop’s motion? Answer in units of kg/s

Answer 1

The numeric value of the coefficient b in the equation R~ = -bv for the air resistance to a raindrop's motion can be calculated using Stokes' law and the given values. The value of b is 0.057 kg/s.

Step-by-step explanation:

The air resistance on a raindrop falling through quiet air is viscous rather than turbulent, so the resistive force depends linearly on the drop's velocity according to the equation

R~ = -bv.

The terminal speed, when the drop stops accelerating, can be calculated using Stokes' law which states that Fs = 6πηrv, where r is the radius of the drop and η is the viscosity of the fluid.

Rearranging the equation gives

v = (mg / (6πηr))

when the force of gravity is balanced by the drag force, and substituting the given values gives

v = (0.36 µg × 9.8 m/s^2) / (6π × (1.00 × 10^-3 kg/m^3) × (2 mm / 2)) = 0.25 m/s.

Since the drop is at terminal velocity, we can equate the two drag equations and solve for b. -bv = 6πηrv.

Rearranging the equation gives

b = (6πηr^2) / v =

(6π × (1.00 × 10^-3 kg/m^3) × (2 mm / 2)^2) / (0.25 m/s)

= 0.057 kg/s.