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

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First you have to look closely at the video and realize that there is not one pinch-off, but two. The first event occurs at the bottom of the neck, near the droplet. After this point, surface tension causes the still-connected neck to begin to assume a spherical form, which pulls the center of the neck upwards. By computing the vertical velocities of every component of the neck, you can determine the overall velocity.

As the neck begins to contract, the overall force pulls the area around the connection point downwards, and eventually the second pinch-off occurs. At this point, knowing the vertical velocities of every part of the resulting blob you can calculate the velocity of the blob.

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    $\begingroup$ Thanks for comment. Yes, 2 droplets and 2 pinch-offs, however, my question is only regarding the small one. I need a clue for calculation of initial upward velocity of the small droplet. I'm not sure what you mean by "knowing the vertical velocities of every part ..." Do you think it's correct to say that droplet upward velocity is the same velocity by which the narrow liquid column is shrinking exactly before the 2nd pinch-off (velocity of the lower head of column)? If yes, how can I theoretically calculate this initial velocity (or at least approximate it)? $\endgroup$ – vorujak Sep 5 '15 at 11:39
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    $\begingroup$ You need to know the surface tension (and probably the viscosity) of the liquid involved. If you know this, and you know the dimensions of the neck just after the first pinch-off occurs, you can calculate the force tending to pull the cylinder (roughly) back up into the parent body, at least until the second pinch-off occurs. The forces will not be simple, since the cylinder will tend to broaden as it gets shorter, and the weight of the blob will tend to pull it away from the parent. $\endgroup$ – WhatRoughBeast Sep 5 '15 at 16:49
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    $\begingroup$ 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? $\endgroup$ – vorujak Sep 5 '15 at 20:57
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    $\begingroup$ 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? $\endgroup$ – vorujak Sep 6 '15 at 1:59
  • $\begingroup$ @WhatRoughBeast Would you please let me know what you think about my calculations in comment above: 1) Are these two forces the only available forces applying to droplet? and 2) Whether the way I'm calculating the surface tension force is correct or not. Thanks $\endgroup$ – vorujak Sep 9 '15 at 22:03

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