Forces at the bottom of a rotating U-tube filled with water

[PSN03]

Homework Statement

Consider a U shaped tube which has enough water so that the horizontal part of the tube is completely filled. The tube now starts rotating about one of the axis. The centripetal force experienced by the water column at the bottom is ?

Relevant Equations

F=mv²/r
v=rw

So when the rotation starts some water will move upwards and in the vertical part of tube.
I know hat centripetal force will be given by
F=mv²/r
Now I though of taking r as centre of mass of the water system but I don't know what to take the value of m as?
Should I only consider the water portion in the horizontal tube or should I take the total water present?

 

[PSN03]

So I did a bit research on the net and found that the force at C will be balanced by the centripetal force and we are only considering the mass in the horizontal part. I don't know why, can anyone explain?

 

[Lnewqban]

For some reason, I can't clearly see the posted picture.

 

[BvU]

Second that.
Additionally: did you render the exercise text exactly as given ?

 

[PSN03]

For some reason, I can't clearly see the posted picture.

I have reuploaded the image. Please do not consider the values given to be related to the problem.

 

[PSN03]

Second that.
Additionally: did you render the exercise text exactly as given ?

The actual problem was something else and I didn't get a particular part of it so I asked that specific part.

The whole question was :

'Length of horizontal arm of a uniform cross section U−tube is l=21 cm and ends of both of the vertical arm are open to surrounding of pressure 10×500 N/m². A liquid of density ρ=10³ kg/m³ is poured into the tube such that liquid just fills the horizontal part of the tube. Now, one of the open ends is sealed and the tube is than rotated about a vertical axis passing through the other vertical arm with angular velocity ω =10 rad/s. If length of each vertical arm is a=6 cm, calculated the length of air column in the sealed arm. (in cm)'

The solution to this problem was:

Due to rotation , let the shift of liquid be X cm
Let the cross sectional area of the tube I.A
In the right limb for compressed ar
P 1V 1 =P 2 V 2
P 1 A×6=P 2 A(6−x)
P 2 = 6P1/6-x

Force at corner C of right limb climb due to liquid above
F1=(P2+dgx)A

It is rotated about left limb. Then the centre petal force is
F 2 =Mω²r=d(l−x)Aω

So my doubt was why is M=d(l-x)A ie the part of the water in the horizontal tube taken into account and why aren't we taking the total mass of the water ie dlA?

PS: later on they equated F1=F2

 

[haruspex]

why is M=d(l-x)A ie the part of the water in the horizontal tube taken into account and why aren't we taking the total mass of the water ie dlA?

It might be more obvious in the rotating frame. The centrifugal acts like a kind of horizontal gravity, so the pressure due to that is related to the 'depth' in the horizontal direction. The fluid in the vertical portion of tube does not contribute to that, just as the fluid in the horizontal portion does not contribute to the pressure caused by gravity acting over the depth in the vertical section.

Whoops - just realized this is misleading. Unlike gravity, the virtual force will vary along the length.

 

[PSN03]

It might be more obvious in the rotating frame. The centrifugal acts like a kind of horizontal gravity, so the pressure due to that is related to the 'depth' in the horizontal direction. The fluid in the vertical portion of tube does not contribute to that, just as the fluid in the horizontal portion does not contribute to the pressure caused by gravity acting over the depth in the vertical section.

Awesome explanation and analogy,Thanks a lot @haruspex.

Will all the liquid in the horizontal tube have the same pressure ie P=P2+dgx during rotation? Cause according to me it should have .

 

[haruspex]

Will all the liquid in the horizontal tube have the same pressure ie P=P2+dgx during rotation? Cause according to me it should have .

No, it will vary with horizontal 'depth' , just as it varies with depth in the vertical section.

See correction to post #7

 

[PSN03]

No, it will vary with horizontal 'depth' , just as it varies with depth in the vertical section.

If that's the case then I can't say that centripetal force on centre of mass would be equal to pressure at C. It should be equal to the pressure difference at the right and the left end of the tube ie P2+dgx-Po , right?
(Where Po is atmospheric pressure as the other end is open)

 

[haruspex]

It should be equal to the pressure difference at the right and the left end of the tube ie P2+dgx-Po , right?

Sorry, I have misled you somewhat. As I have now clarified in post #7, it's not quite the same as gravity since the acceleration varies along the length. But this principle still stands: at the junction, from the vertical section we have the pressure of the trapped air plus the pressure from gravity on the fluid in that section, while from the horizontal section we have atmospheric pressure plus the pressure from the centrifugal force acting on the horizontal fluid; and these two pressures must balance.
So your original approach was correct, and I was only intending to clear up your doubt.

That said, I didn't understand this step:

Mω²r=d(l−x)Aω

d is density, yes? What happened to r and the other ω?

 

[PSN03]

Sorry, I have misled you somewhat. As I have now clarified in post #7, it's not quite the same as gravity since the acceleration varies along the length. But this principle still stands: at the junction, from the vertical section we have the pressure of the trapped air plus the pressure from gravity on the fluid in that section, while from the horizontal section we have atmospheric pressure plus the pressure from the centrifugal force acting on the horizontal fluid; and these two pressures must balance.
So your original approach was correct, and I was only intending to clear up your doubt.

That said, I didn't understand this step:

d is density, yes? What happened to r and the other ω?

But according to the solution that I have, they haven't considered the atmospheric pressure in the left part while equating the forces, so is the solution provided wrong?

And yes d is density,
r=distance of centre of mass from point of rotation=l+x/2
Therefore
Mω²r=d*A*(l-x)*ω²*(l+x)/2
(Sorry I did a mistake before)

 

[haruspex]

is the solution provided wrong?

Try a sanity check: what do you get for x with ω=0?

 

[PSN03]

Try a sanity check: what do you get for x with ω=0?

It should be zero as no rotation is happening. The initial condition is same.

 

[haruspex]

It should be zero as no rotation is happening. The initial condition is same.

Right, but is that what the book answer gives? Yours?

 

[PSN03]

Right, but is that what the book answer gives? Yours?

Yes, initially there is no rotation so the water in only in the horizontal tube with x=0
Once the rotation starts we find some water in the vertical tube as well with its length as x. The two diagrams also say so.

 

[PSN03]

I am not getting the point that why aren't they taking atmospheric pressure into consideration with the centrifugal force.

 

[haruspex]

I am not getting the point that why aren't they taking atmospheric pressure into consideration with the centrifugal force.

I have not seen their detailed calculation. Perhaps they calculated the pressure in the compressed region as only its increase in pressure.

 

[PSN03]

I have not seen their detailed calculation. Perhaps they calculated the pressure in the compressed region as only its increase in pressure.

The solution provided by them is:

 

[haruspex]

The solution provided by them is:

Hard to decipher because of all the typos. δ=6? ρ₀=P1?
But it seems to me to fail my sanity check. Did you plug ω=0 into it as I suggested?

 

[PSN03]

Hard to decipher because of all the typos. δ=6? ρ₀=P1?
But it seems to me to fail my sanity check. Did you do plug ω=0 into it as I suggested?

I think I understood what my mistake was.
Thanks for your help and time, means a lot to me