Calculating Vorticity of 2-D Flow Motion

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SUMMARY

The discussion focuses on calculating the vorticity of a 2-D flow motion defined by the velocity field (u,v) = (y,-x). The vorticity is calculated using the formula \(\omega = v_x - u_y\), leading to a final result of \(\omega = 2\). The confusion arises from the initial calculation of \(\omega = -2\), which is incorrect. The correct application of the vorticity definition, \(\vec{\omega}=\vec{k}\Big(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\Big)\), clarifies the calculation process.

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  • Understanding of vector calculus, specifically curl and divergence
  • Familiarity with fluid dynamics concepts, particularly vorticity
  • Knowledge of partial derivatives in the context of 2-D functions
  • Basic proficiency in mathematical notation and equations
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squenshl
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Homework Statement


Consider 2-D flow motion (u,v) = (y,-x). Calculate the vorticity field of the flow.


Homework Equations


\omega = vx - uy


The Attempt at a Solution


I calculated \omega = -2, so using the vorticity equation I get zero. So I guess what I am asking is what exactly am I calculating.
 
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Vorticity is defined as \mathbf{\omega}=\nabla\times\mathbf{u}, as far as I can remember the vorticity equation describes the change in vorticity.
 
squenshl said:
I calculated \omega = -2, so using the vorticity equation I get zero.

What do you mean by this?

Vorticity is a vector field, defined as the curl of the fluid velocity. For a 2D flow this reduces to
\vec{\omega}=\vec{k}\Big(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\Big)

What do you get when you use this definition?
 
The answer is \omega = 2
 
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?

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