Understanding Angular Velocity and Curl for a Rotating Fluid

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A fluid rotating about the z-axis exhibits angular velocity, which can be expressed using the right-hand screw rule. The velocity of a point in the fluid is determined by the angular velocity and the radius, with the relationship given by the cross product of the angular velocity vector and the position vector. For constant angular velocity, the curl of the velocity field results in a value of 2w_k, while if angular velocity varies inversely with the square of the radius, the curl becomes zero. The discussion emphasizes the ability to calculate the curl in both Cartesian and polar coordinates without relying on Stokes' theorem. Understanding these relationships is crucial for analyzing fluid dynamics in rotating systems.
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A fluid rotates with an angular velocity w about the z-axis. The direction of rotation is related to the z-axis by the rigrt hand screw rule.
a) Find the velocity v of a point in the fluid, and show that
curl v = 2w_k

b) If now w is a function of the radius r, show that curl v = zero if w= constant/r square
 
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First show us what you have done! Certainly if your teacher has asked you to do this, you must to how to find the curl of a vector function- what is the formula?
 
what u are asking to find out is vorticity(curl of v). isn't it?

vorticity=U(say)

U= del X v...(i)

find out the cross product using matrix and separate i,j and k terms...

take divergence of U but since divergence of curl =0 u will have

del.U =0

use stokes theorem to write the integral form of eqn (i)

vel in theta direction i,e v(theta)=wo r...where w0 can be the angular vel of any point on the fluid and vel along vector r direction i,e vr=0

find out delXv in polar coordinates and put v(theta)=wo r and vr=0...[note here theta,o and r (for vr)are suffixes]

u will get the ans 2wk[reqd z component,i guess k is unit vector in z direction]

I think it will be like this
 
You still need to find the expression for v, and if you do that you can simply find the curl of v directly using del X v in cartesian coordinates as well as polar.
 
OlderDan said:
You still need to find the expression for v, and if you do that you can simply find the curl of v directly using del X v in cartesian coordinates as well as polar.

question is to find the curl v and not v...what you have in mind?
 
aura said:
question is to find the curl v and not v...what you have in mind?

v is determined by w and r in both cases. You wrote two components of v in terms of w in cylindrical coordinates. You can write v in cartesian coordinates if you choose. However you write it, you can perform the operation del X v without using Stokes theorem or doing any integrals.
 
OlderDan said:
v is determined by w and r in both cases. You wrote two components of v in terms of w in cylindrical coordinates. You can write v in cartesian coordinates if you choose. However you write it, you can perform the operation del X v without using Stokes theorem or doing any integrals.


I don't understand. please explain to me again. Thank you.
 
danai_pa said:
I don't understand. please explain to me again. Thank you.

The problem states that the fluid rotates about the z axis. In the first part, the angular velocity is constant. In the second part the angular velocity is inversely proportional to the square of the distance r from the z axis. In either case, a particle of water is in circular motion about the axis with velocity that depends on where it is in relation to the axis. For rotation about an axis, you can find the velocity if you know the angular velocity and the radius of rotation. In terms of vectors, the velocity under these conditions is

\vec v = \vec \omega \times \vec r

In the first case

\vec \omega = \omega_k\ \widehat k

In the second case

\vec \omega = \frac {\omega_0}{r^2}\ \widehat k

In both cases

\vec r = x\ \widehat i +y\ \widehat j +z\ \widehat k

The cross product is going to yield velocities with components parallel to the x-y plane, independent of the z coordinate, circulating about the z-axis. In the first case, velocity will increase with distance from the z-axis. In the second case, velocity will decrease with distance from the z-axis. Remembering that the distance r from the z-axis can be written as a function of x and y, you can find the curl of v

\nabla \times \vec v

in Cartesian coordinates. If you prefer, you can write the velocity in cylindrical coordinates and do the problem that way.
 
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