6.5.6 Advanced Topic: Stopping Time and Distance

Suppose a small spherical object of mass m and radius r is moving at its terminal speed $v_{term}$ when the applied force suddenly vanishes, leaving only drag. How long does it take the object to stop, and how far does it move while stopping? We cannot use constant-acceleration kinematics to answer these questions, so we need to use the more general idea of instantaneous acceleration that we introduced in 《 SECTION 2.7.

We will assume that the object is moving to the right, so the x-components of the drag force and the acceleration are negative. From Stokes’ law, the acceleration is

a_{x} = \frac{(F_{drag})_{x}}{m} = - \frac{6 π η r v_{x}}{m}

(6.20)

By definition, the object’s acceleration is $a_{x} = d v_{x} / d t$ . Thus the motion is described by the equation

\frac{d v_{x}}{d t} = - \frac{6 π η r v_{x}}{m}

(6.21)

This looks complicated, but we can find the velocity by integration.

First, we rearrange Equation 6.20 to get all the velocity terms on one side:

\frac{d v_{x}}{v_{x}} = - \frac{6 π η r}{m} d t

The fraction on the right is a group of constants, so it is constant. Now we integrate both sides from an initial speed $v_{term}$ at time $t_{0} = 0$ to a later speed v at time t:

\int_{v_{term}}^{v} \frac{d v_{x}}{v_{x}} = - \frac{6 π η r}{m} \int_{0}^{t} d t = - \frac{6 π η r}{m} t = - \frac{t}{τ}

(6.22)

In Equation 6.22 we took the constants outside the time integration. Now we can group them together and define the time constant $τ$ (Greek tau) of the motion as

τ = \frac{m}{6 π η r}

(6.23)

It’s not obvious, but by using the units of viscosity $(Pa s)$ we can show that the units of $τ$ are s, as they should be for a time. We’ll see that the time constant is a characteristic time for how long it takes the object to come to a halt.

To finish the integration, recall from calculus that $\int d x / x = ln x, w h e r e In x$ is the natural logarithm of x. In this case,

\int_{v_{term}}^{v} \frac{d v_{x}}{v_{x}} = ln v_{x} | \begin{array}{l} v \\ v_{term} \end{array} = ln v - ln v_{term} = ln (\frac{v}{v_{term}})

(6.24)

Thus

ln (\frac{v}{v_{term}}) = - \frac{t}{τ}

(6.25)

To solve for v, we also need to recall that the logarithmic function and the exponential function are inverses of each other, which means that $ln (e^{x}) = x$ and $e^{ln x} = x$ . Applying the latter of these to Equation 6.25, we find

e^{ln (v / v_{term})} = \frac{v}{v_{term}} = e^{- t / τ}

(6.26)

If we multiply through by $v_{term}$ , we find that the object’s speed decreases exponentially with time:

v = v_{term} e^{- t / τ}

(6.27)

Equation 6.27 for the object’s speed, which is graphed in FIGURE 6.23, is called an exponential decay. It is a behavior that will reappear several times throughout this text. The graph starts from $v (t = 0) = v_{term}$ because $e^{0} = 1$ . It then asymptotically approaches zero for very large values of t.

The figure illustrates the velocity versus time graph for an object coasting to a halt at low Reynold's number.

FIGURE 6.23 Full Alternative Text

When $t = τ$ , $v = e^{- 1} v_{term} = 0.37 v_{term}$ . When $t = 2 τ$ , $v = e^{- 2} v_{term} = 0.13 v_{term}$ . For any quantity that decays exponentially, the quantity decays to 37% of its initial value after one time constant has elapsed. Exponential decay is a common mathematical model; once you’ve learned the properties of exponential functions, you will immediately know how to apply this knowledge to a new situation.

As a practical matter, we could say that the motion has pretty well ceased at $t = 2 τ$ , so we define a stopping time

Δ t_{stop} = 2 τ = \frac{m}{3 π η r}

(6.28)

We’ll omit the details, but an integration of $v_{x} = d x / d t = v_{term} exp (- t / τ)$ would find that the total distance the object traveled while stopping is

Δ x_{stop} = v_{term} τ = \frac{1}{2} v_{term} Δ t_{stop}

(6.29)

EXAMPLE 6.10 A Paramecium coasts to a stop

Paramecia, common unicellular inhabitants of freshwater ponds, are about $100 μ m$ in size and swim at about $200 μ m/s$ . If a Paramecium in 20°C water suddenly stops swimming, how long does it take to come to a stop and how far does it coast while stopping?

MODEL A Paramecium’s swimming speed is its terminal speed $v_{term}$ in response to the pushing force generated by beating its cilia. Although Paramecia are somewhat elongated, we’ll simplify the situation by modeling the Paramecium as a $100 -μ m-diameter$ sphere. The density of a Paramecium is essentially that of water, so we can calculate its mass from the density of water and the volume of a sphere.

SOLVE The mass is $m = \frac{4}{3} π r^{3} ρ$ , where $ρ$ is the density of water. If we use this expression for the Paramecium’s mass in Equation 6.28, then the time it takes to come to a stop is

\begin{matrix} Δ t_{stop} & = & \frac{\frac{4}{3} π r^{3} ρ}{3 π η r} = \frac{4 r^{2} ρ}{9 η} & = & \frac{4 (50 \times 10^{- 6} {m)}^{2} (1000 {kg/m}^{3})}{9 (1.0 \times 10^{- 3} Pa s)} \\ = & 1.1 \times 10^{- 3} s \approx 1 ms \end{matrix}

The density and viscosity of water are taken from Table 6.2. During this time, the Paramecium travels distance

\begin{matrix} Δ x_{stop} & = & \frac{1}{2} v_{term} Δ t_{stop} = \frac{1}{2} (200 \times 10^{- 6} m/s)(1.1 \times 10^{- 3} s) \\ = & 1.1 \times 10^{- 7} m \approx 0.1 μ m \end{matrix}

REVIEW Stopping in 1 ms over a distance of $0.1 μ m,$ which is only $\frac{1}{1000}$ the diameter of the organism, is, for all practical purposes, an instantaneous stop. This agrees with our assertion about motion at low Reynolds number.