Inertial pressure on in a rotating syringe

First question on stackexchange. Hopefully somebody can help me out. I'm struggling to express the following eloquently.

I have a rotating syringe tipped with a needle. rotation speed w, with radius r1 and r2.

Now let us pretend that the needle is blocked, so only r1 matters.

My thinking is, due to inertia the pressure observed at r1 is equal to density x fluid 'height' (in an horizontal sense) x acceleration due to inertia.

I'm struggling to calculate this outward acceleration. I've always been told that centrifugal force doesn't exist. So is inertia equal to the centripetal acceleration? It isn't made entirely clear...

Many thanks in advance, if that is too easy, you could also consider the case where the needle is blocked just at the tip (r2) what would be the pressure there.

• I have put on a homework and exercises tag as this question is in that category. – tom Nov 29 '14 at 15:28
• I've always been told that centrifugal force doesn't exist If it would not exist, then there would be no reason for your laundry to become dry in your washing machine. – Agnius Vasiliauskas Dec 12 '20 at 11:56
• @AgniusVasiliauskas From the perspective of an observer standing next to the washing machine, the water is separated from the laundry exactly because there is no force on the water. – Filippo Jan 14 at 8:51
• From the perspective of a rotating observer inside the washing machine, a force acts on the water. – Filippo Jan 14 at 8:53
• So, seems, that you agree that indeed centrifugal force exists in a non-inertial reference frames. Thus your comment in a post to which I replied is still not valid, cause it implies that centrifugal force does not exist in any reference frames at all :-) – Agnius Vasiliauskas Jan 14 at 21:21

The way I see it this problem is similar the problem of column of water and the pressure rising due to gravity pulling down on the liquid - in which case pressure, $P$, is given by

$P = h \rho g$

where $h$ is the height of liquid, $\rho$ is the density and $g$ the acceleration due to gravity - but you probably knew this already.

The point I want to make here is that the height of water is important because the forces all add up from top down to bottom (so the pressure at the bottom of the oceans can be very very high - 100s or 1000s or more atmospheres)

Here the issue is the that the force (and hence acceleration) due to the rotation changes - as the radius gets larger the force gets bigger of course.

So calculate the pressure at $r1$ and $r2$ you need to do an integration to add up all the forces for the liquid 'above' $r1$ and $r2$ - so on your diagram you need the height (edit - ah not labelled, but that it is in writing on the diagram.)

The good news is that the formulae for $r1$ and $r2$ will be the same as the cross sectional area of the tube does not matter - but again you need to know the height of liquid - or length of liquid towards the centre.

Finally, Centripetal force = centrifugal - action and reaction... here you can consider the centrepetal force is the same as the force the centrifugal force. Think of it this way, the pressure at $r1$ or $r2$ must provide the inward centrepetal force on the column of liquid above it to make it move in circular motion.