52.8: Stokes’s Law
- Page ID
- 92334
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Stokes's law gives the resistive force on a sphere moving through a viscous fluid. It was developed by the 19th century English physicist and mathematician George Stokes. Stokes's law states that the resistive force on the sphere is given by
\[F_R=6 \pi \mu r v, \]
where \(F_R\) is the resistive force on the sphere, \(r\) is its radius, \(\mu\) is the dynamic viscosity of the fluid, and \(v\) is the relative velocity between the fluid and the sphere. This is generally valid for low Reynolds numbers ( \(\operatorname{Re}<1\) ).
Notice that the Stokes's law force is of the form of a Model I resistive force described in Chapter 22 ( \(F_R \propto v\) ), with the resistance coefficient \(b=6 \pi \mu r\). By Eq. 22.1.21 the terminal velocity for Model I is \(v_{\infty}=m g / b\); so for a sphere moving through a viscous fluid, we have by Stokes's law
\[v_{\infty}=\frac{m g}{6 \pi \mu r} . \]
What is the terminal velocity of a steel ball of diameter 1 cm falling through a jar of honey? Solution. Taking the density of steel as \(\rho=7.86 \mathrm{~g} / \mathrm{cm}^3\), we find the mass of the steel ball as
\[m=\rho V=\rho\left(\frac{4}{3} \pi r^3\right)=4.115 \mathrm{~g}=4.115 \times 10^{-3} \mathrm{~kg} .\]
Solution
From Table 52.6.1, the dynamic viscosity \(\mu\) of honey is 5 Pa s ; the terminal velocity is then given by Eq. \(\PageIndex{2}\):
\[
\begin{aligned}
v_{\infty} & =\frac{m g}{6 \pi \mu r} \\
& =\frac{\left(4.115 \times 10^{-3} \mathrm{~kg}\right)\left(9.80 \mathrm{~m} / \mathrm{s}^2\right)}{6 \pi(5 \mathrm{~Pa} \mathrm{~s})\left(0.5 \times 10^{-2} \mathrm{~m}\right)} \\
& =8.56 \mathrm{~cm} / \mathrm{s} .
\end{aligned}
\]