Electro Dynamics Curl Question (semi-basic)

[steve233]

Homework Statement

(I use \x for a vector)

A fluid has an angular velocity u about the z-axis (right hand rule). the velocity of a point in the fluid is
\v = \u X \r.

\u = u₀ / s² (z-hat). (u-naught is a constant)
Find \DEL X \v (del as in the symbol del)

Homework Equations

The curl in cylindrical coordinates.

The Attempt at a Solution

First I find \v by taking the cross product of \u and \r which turns out to be:
u₀ / s (theta hat)

Then using the equation of the curl in cylindrical coordinates (only the last part of it since the rest turns out to be 0), I end up getting 0 (z-hat) which can't possibly be the correct answer. I'm missing one small piece and I'm not too sure what it could be.

If the question is unclear, I can reword it better.
Thanks in advance.

 

[anathema]

Hello,

Thank you for posting your question on the forum. It seems like you are on the right track with your solution. However, there are a few mistakes that I would like to point out. Firstly, the velocity of a point in the fluid is given by \v = \u X \r, which is incorrect. The correct expression should be \v = \u X \r, where \u is the angular velocity and \r is the position vector of the point in the fluid. Also, the given expression for \u is incorrect. It should be \u = u0 / s^2 (z-hat), where s is the distance from the z-axis to the point in the fluid and u0 is a constant.

Now, to find \DEL X \v, we first need to express \v in cylindrical coordinates. Substituting the given values, we get \v = u0 / s (theta hat). Now, using the expression for the curl in cylindrical coordinates, we get \DEL X \v = (1/s) (d/ds (s * u0 / s)) = u0 / s^2 (z-hat). This is the correct answer.

I hope this helps. If you have any further questions, please feel free to ask.