Two rain drops each of radius r are falling with terminal velocity vt while in motion they coalesce into one drop what is the terminal velocity of the drop so formed?

1 Answer
Jun 14, 2018

#v'_t = 2^(1/6) v_t#

Explanation:

For Rayleigh Drag:

  • #F_D = - 1/2 rho v^2A C_d#

  • #rho# is fluid density, #A# is cross sectional area of object orthogonal to motion, and #C_d# is a constant

A falling object will accelerate as:

  • # F =ma=mg-1/2 rho v^2A C_d#

At terminal velocity, #a = 0#, leading to this expression for terminal velocity:

# v_t= sqrt ((2mg)/( rho AC_d)) #

Taking out the non-constants:

  • # v_t propto sqrt ( m / A) qquad square#

If two drops coalesce to form a new sphere, conservation of mass:

  • #m' = 2m#

For new radius #r'#:

  • # 4/3 pi (r')^3 = 2 * 4/3 pi r^3 #

  • # implies bb( r^' = 2^(1/3) r )#

  • #A' = pi (r^')^2 = 2^(2/3) A #

  • #(m')/(A') = 2^(1/3) m/A#

From #square#:

#(v'_t)/v_t = 2^(1/6) #