Flux Integral of a Fluid Rotating about an Axis

Click For Summary
SUMMARY

The discussion centers on determining the placement of a unit square, S, in the yz-plane to achieve zero net fluid flow through it, given a fluid with density ρ rotating about the z-axis with angular velocity ω. The velocity vector is derived from the cross product of angular velocity and position, represented as v = ω × r. The consensus is that it is impossible to position the square such that there is no net fluid flow, as the center of the square must be located in a restricted subset of the yz-plane to balance the fluid flow in the positive and negative x-directions.

PREREQUISITES
  • Understanding of fluid dynamics principles, specifically fluid rotation.
  • Familiarity with vector calculus, including cross products.
  • Knowledge of angular velocity and its implications in fluid motion.
  • Ability to interpret and apply integral equations in physics.
NEXT STEPS
  • Study the implications of angular momentum in rotating fluids.
  • Learn about the application of the divergence theorem in fluid dynamics.
  • Explore the concept of vorticity and its role in fluid flow analysis.
  • Investigate the mathematical modeling of fluid flow using Navier-Stokes equations.
USEFUL FOR

Students and professionals in physics, particularly those focusing on fluid dynamics, mechanical engineers, and anyone involved in the study of rotational motion in fluids.

vs74043
Messages
2
Reaction score
0

Homework Statement


We have a fluid with density ρ which is rotating about the z-axis with angular velocity ω. Where should a unit square, call it S, be placed in the yz-plane such that there is zero net amount of fluid flowing through it?

Homework Equations


$$\mathbf{v}=\mathbf{\omega}\times\mathbf{r}$$
$$\int\mathbf{F}\cdot d\mathbf{S}$$

The Attempt at a Solution


My attempt has been to try and show that the velocity vector (cross product of angular velocity and position) is independent of z. This would mean that I can position the square such that one side is on the y-axis at distance δ from the origin. Then just find the value of δ.

I'm not sure how to set-up my angular velocity vector and my position vector. Also, is my method idea correct?
 
Physics news on Phys.org
vs74043 said:
that there is zero net amount of fluid flowing through it
There is nowhere you can put it that there is no fluid flowing through it. So focus on that word I have bolded. If the square is in the yz plane, the centre of the square can only be in a very restricted subset of locations for there to be as much fluid flowing through in the positive x direction as there is in the negative x direction.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 0 ·
Replies
0
Views
1K
Replies
19
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K