# Why small droplet goes upward after pinch-off?

This small droplet moves against gravity. How to calculate its initial upward velocity exactly after pinch-off?

• Based on figure above, I assumed that we have two sources of force and performed the following calculations: $F=\pi.d.\sigma$ and $W=m.g=\rho.g.\pi.D^3/6$ After replacing $D=0.4 mm$ and $d=0.05 mm$ and using water as liquid we have: $F=1.15e-5 N$ and $W=0.03e-5N$ Therefore, $\Delta F=1.12e-5=m.a$ ==> $a=300 m/s^2$ I know that the amount of upward force is changing as the neck diameter is reducing and so I assumed an average $d$, but this huge acceleration is far from even an approximate answer and so I think something is not correct here. What do you think? – vorujak Sep 5 '15 at 20:57
• There is also a force which is applied to the stretched liquid column (left figure) to retract it to a spherical shape to maintain the minimum energy. I'm wondering whether this force is considered in $F=\pi.d.\sigma$ or you should calculate it separately. Does anybody know this for sure? – vorujak Sep 6 '15 at 1:59