5.14 Terminal Velocity
Consider a water droplet having radius r such as that of fog falling vertically, the air drag on the water droplet increases with speed. The droplet accelerates rapidly under the over powering force of gravity which pulls the droplet rapidly downward due to force of gravity. However, the upward drag force on it increases as the speed of the droplet increases. The net force on the droplet is:
Net force=Weight−Drag force…(5.27)
As the speed of the droplet continues to increase, the drag force eventually approaches the weight in the magnitude. Finally, when the magnitude of the drag force becomes equal to the weight, the net force acting on the droplet is zero. Then the droplet will fall with constant speed called terminal velocity.
To find the terminal velocity vt, in this case, we use Stokes' law for the drag force. Equating it to the weight of the drop, we have:
0=mg−6πηrvt
vt=mg/(6πηr)…(5.28)
The mass of the droplet is ρV, where V=34πr3 is the volume of the sphere.
Substituting above values in the Eq. (5.28), we have:
vt=9η2gr2ρ…(5.29)
Example 5.5
A tiny water droplet of radius 0.010cm descends through air from a high building. Calculate its terminal velocity. Given that η for air =19×10−6kgm−1s−1 and density of water ρ=1000kgm−3.
Solution:
r=1.0×10−4m, ρ=1000kgm−3, η=19×10−6kgm−1s−1
Putting the above values in Eq. (5.29):
vt=[9×19×10−6kgm−1s−1][2×9.8ms−2×(1.0×10−4m)2×1000kgm−3]
Terminal velocity=1.1ms−1