When an object moves through a fluid, the fluid exerts a viscous force on the ob

Question

When an object moves through a fluid, the fluid exerts a viscous force on the object that tends to slow it down. For a small sphere of radius R, moving slowly with a speed v, the magnitude of the viscous force is given by Stokes' law, F = 6p?Rv, where ? is the viscosity of the fluid.
(a) What is the viscous force on a sphere of radius R = 5.25 10-4 m that is falling through water (? = 1.00 10-3 Pa · s) when the sphere has a speed of 3.3 m/s?
N

(b) The speed of the falling sphere increases until the viscous force balances the weight of the sphere. Thereafter, no net force acts on the sphere, and it falls with a constant speed called the "terminal speed." If the sphere has a mass of 9.90 10-6 kg, what is its terminal speed?
m/s

Explanation / Answer

Hints: (a) We know that F = 6 * pi * n * R * v Given that pi = 3.14 n = 1.00 x 10^-3 kg /m -s v = 3.3 m/s R = 5.25 x 10^-4 m Substitute values and solve for F. (b) Vs = [ 2(Pp - Pf) g R^2 ] / (9 n) Where g is the gravitational acceleration (m/s2), ?p is the mass density of the particles (kg/m3), and ?f is the mass density of the fluid (kg/m3). Substitute values and solve for Vs.

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