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Physics · Ch 2 — Mechanical Properties of Fluids

Terminal Velocity

2.7.1

Terminal Velocity

Consider a spherical object falling under gravity through a viscous fluid. Three separate forces act on it during this downward fall: (1) the viscous force FvF_v, directed upward, whose magnitude keeps increasing as the object's downward velocity increases; (2) the gravitational force, or weight, FgF_g, directed downward, and constant throughout; and (3) the buoyant force, or upthrust, FuF_u, directed upward, and also constant throughout.

The net downward force is f=Fg−(Fv+Fu)f = F_g - (F_v + F_u), and it is this net force that is responsible for the object's initial increase in speed as it starts falling. Since FgF_g and FuF_u stay fixed while FvF_v keeps growing as the speed builds up, a stage is eventually reached at which the net force f becomes exactly zero, i.e. Fg=Fv+FuF_g = F_v + F_u. From that point onward, the object's downward velocity remains constant — this constant downward velocity is called the terminal velocity, and, since the speed no longer changes beyond this point, the viscous force FvF_v itself also stays constant from here on. (This entire discussion necessarily applies only to a streamline flow around the falling object.)

Let the falling sphere have radius r, mass m and density ρ, and let the surrounding medium have density σ and coefficient of viscosity η. Once the sphere has reached its terminal velocity, the total downward force on it exactly balances the total upward force:

weight of sphere (mg)=viscous force+buoyant force due to the medium\text{weight of sphere } (mg) = \text{viscous force} + \text{buoyant force due to the medium}

43πr3ρg=6πηrv+43πr3σg\dfrac{4}{3}\pi r^3 \rho g = 6\pi\eta r v + \dfrac{4}{3}\pi r^3 \sigma g

6πηrv=43πr3g(ρ−σ)6\pi\eta r v = \dfrac{4}{3}\pi r^3 g(\rho - \sigma)

v=2r2(ρ−σ)g9η— (2.37)\boxed{v = \dfrac{2r^2(\rho - \sigma)g}{9\eta}} \qquad \text{--- (2.37)}

This is the expression for the terminal velocity of the falling sphere. From Eq. (2.37), the same relation can equally be rearranged to give the coefficient of viscosity of the medium:

η=2r2(ρ−σ)g9v— (2.38)\eta = \dfrac{2r^2(\rho - \sigma)g}{9v} \qquad \text{--- (2.38)}

— a form especially useful for measuring a fluid's coefficient of viscosity experimentally, simply by dropping a sphere of known radius and density into it and timing its terminal (constant) fall speed. …

Figure 2.32Fig. 2.32: Forces acting on an object moving through a viscous medium — upthrust F_u plus viscous force F_v acting upward and the weight mg (F_g) downward on a sphere in a fluid
Fig. 2.32 — Fig. 2.32: Forces acting on an object moving through a viscous medium — upthrust F_u plus viscous force F_v acting upward and the weight mg (F_g) downward on a sphere in a fluid

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. A sphere of radius r, mass m and density ρ is shown falling through a surrounding fluid medium of density σ and coefficient of viscosity η, with three labelled force arrows: the weight FgF_g pointing straight down, and both the viscous drag force FvF_v and the buoyant force (upthrust) FuF_u pointing straight up — together forming the three-force balance, Fg=Fv+FuF_g = F_v + F_u, that holds once the sphere reaches its terminal (consta …

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