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Hydroynamics: Stoke's Law, Find work done by drag force

  1. Nov 12, 2012 #1
    1. The problem statement, all variables and given/known data
    A water droplet starts falling from rest from a height [itex]h[/itex] and a quires the terminal speed just before reaching the ground. If [itex]g[/itex](acceleration due to gravity) is assumed to be constant and the radius of the drop is [itex]r[/itex], find the work done by air drag. The density of air is [itex]\rho_{\text{air}}[/itex] and the density of water is [itex]\rho_{\text{W}}[/itex]


    2. Relevant equations
    Stoke's Law: [itex]F_{\text{drag}}=6\pi \eta r v[/itex]


    3. The attempt at a solution
    Let [itex]a[/itex] be the acceleration of the droplet. Then,

    [tex]\begin{aligned}mg-F_{\text{upthrust}}-F_{\text{drag}} &= ma \\ ma &= mg-mg(\frac{\rho_{\text{air}}}{\rho_{\text{W}}}) -6\pi \eta rv \\ a &= g\left(1-\frac{\rho_{\text{air}}}{\rho_{\text{W}}}\right)-\frac{6\pi \eta r }{m}v \\ v\frac{dv}{ds} &=g\left(1-\frac{\rho_{\text{air}}}{\rho_{\text{W}}}\right)-\frac{6\pi \eta r }{m}v \\ ds &=\frac{v}{g\left(1-\frac{\rho_{\text{air}}}{\rho_{\text{W}}}\right)-\frac{6\pi \eta r }{m}v } dv \end{aligned}[/tex]

    The work done by the drag force is

    [tex]\begin{aligned} dW &= -F_{\text{drag}} ds \\ &= -(6\pi \eta r v)\left[ \frac{v}{g\left(1-\frac{\rho_{\text{air}}}{\rho_{\text{W}}}\right)-\frac{6\pi \eta r }{m}v } dv\right] \\ \end{aligned}[/tex]

    Am I doing correct? The answer given in my book is

    [tex]\frac{4}{3}\pi r^3\rho_{\text{W}}g\left[ \frac{2r^4 g}{81\eta^2}(\rho_{\text{W}}-\rho_{\text{air}})^2-h\left( 1- \frac{\rho_{\text{air}}}{\rho_{\text{W}}}\right) \right][/tex]
     
  2. jcsd
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