Chapter 5
Modeling with Differential Equations
5.1 Raindrops
5.1.2 Stokes' Law
For very small spherical droplets (those of diameter inches) falling in still air, experience suggests that the drag force of the air (thus also the corresponding component of acceleration) is proportional to the velocity of the drop. Specifically, this model (called Stokes' Law, after G. G. Stokes) is expressed in the form of a differential equation by
Experimental evidence suggests that the constant \(k\) is approximately
when the units are feet and seconds.
The differential equation for Stokes' Law implies the existence of a terminal velocity, a speed that the drops approach but cannot exceed.
Calculate the terminal velocity, , for drops small enough to obey Stokes' Law. In particular, what is the terminal velocity of a drizzle drop with inches ?
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We will see shortly that drops this size reach their terminal velocities very quickly, so it is a reasonable approximation to assume that such drops are traveling at terminal velocity for the entire fall. With that assumption, how long does it take for a drizzle drop to fall \(3000\) feet?
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Stokes' Law, together with the assumption of zero starting velocity, constitutes the initial value problem
\(\frac{dv}{dt}=g-kv\) with \(v(0)=0\).Rewrite the differential equation
\(\frac{dv}{dt}=g-kv\)in the form
\(\frac{dv}{dt}=-k\left(v-\frac{g}{k}\right)\)Make the substitution \(y=v-\frac{g}{k}\), and determine the corresponding initial value problem for \(y\).
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Solve the initial value problem from (a) for \(y\). Then substitute back to obtain a formula for \(v\) as a function of \(t\).
How does your formula confirm your calculation of the terminal velocity in Activity 1?
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Use the results from Activities 1 and 2 to find a formula for the ratio
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How long does it take a drizzle drop to reach \(99\)% of terminal velocity?
Was it reasonable to assume that these drops travel at terminal velocity for their entire fall? Why or why not?