Shape of a bubble inside a rotating container

Raihan amin
Messages
45
Reaction score
0
1.

A closed cylindrical vessel filled with water (at room temperature) contains a small air bubble of normal pressure and volume ##V=1~{cm}^3## inside in it.The cylinder is then started to be rotated slowly with a small angular acceleration in a complete weightlessness (at a space station) .When the angular velocity of the cylinder has reached to ##w_0=30~s^{-1}##,it is kept constant.
Find the stationary shape of the air bubble.The surface tension of the water is ##\sigma=0.7~ N/m##.
The extra pressure inside a bubble is related to the surface tension is as follows:

$$∆P=\frac{2\sigma}{R}$$.
And for cylindrical symmetry, the pressure inside the container will be
$$P=\frac{1}{2}{\rho}w^2r^2$$

From V we can find ##R_0##.
But i can't figure out how the velocity of water particle will affect the shape of the bubble and what is the advantage of weightlessness?
Thanks.

 
Last edited:
Physics news on Phys.org
The advantage of weightlessness is that you don't have to worry about the direction of the axis of rotation relative to the direction of gravity. Space is isotropic.
Raihan amin said:
But i can't figure out how the velocity of water particle will affect the shape of the bubble ...
In weightlessness, if the cylinder is not rotated, the bubble will be spherical.
Question 1: Will the bubble still be spherical when the cylinder starts rotating in weightlessness? Why or why not?
Question 2: If the bubble is at some arbitrary point (r, θ, z) at the moment the cylinder starts rotating, will it stay there? If yes, what will its subsequent motion be? If no, why not?
 
  • Like
Likes Raihan amin
kuruman said:
Question 2: If the bubble is at some arbitrary point (r, θ, z) at the moment the cylinder starts rotating, will it stay there? If yes, what will its subsequent motion be? If no, why not?
As the cylinder is closed,we can consider it, along with the water,as a rigid body.Then we can say that the bubble will keep it's position fixed.
 
And i have no idea about the shape of the bubble when the cylinder starts rotating. Please give the actual answers of your questions .
 
What would the pressure distribution be inside the rotating container if the bubble were not present?
 
Raihan amin said:
And i have no idea about the shape of the bubble when the cylinder starts rotating. Please give the actual answers of your questions .
This is how it works: You ask a question, we lead you to the answer by asking intermediate questions that you must answer yourself.
Raihan amin said:
As the cylinder is closed,we can consider it, along with the water,as a rigid body.Then we can say that the bubble will keep it's position fixed.
Shouldn't the rotation make a difference? What about artificial gravity?
https://en.wikipedia.org/wiki/Artificial_gravity
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top