Understanding Angular Velocity and Curl for a Rotating Fluid

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In summary: The result will be the same.In summary, the problem deals with a fluid rotating about the z-axis with a constant angular velocity in the first part and a variable angular velocity in the second part. To find the velocity of a point in the fluid, the formula \vec v = \vec \omega \times \vec r is used. The resulting velocity vector is then used to find the curl of v, which can be done in either Cartesian or cylindrical coordinates.
  • #1
danai_pa
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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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  • #2
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?
 
  • #3
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
 
  • #4
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.
 
  • #5
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?
 
  • #6
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.
 
  • #7
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.
 
  • #8
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

[tex] \vec v = \vec \omega \times \vec r [/tex]

In the first case

[tex] \vec \omega = \omega_k\ \widehat k [/tex]

In the second case

[tex] \vec \omega = \frac {\omega_0}{r^2}\ \widehat k [/tex]

In both cases

[tex] \vec r = x\ \widehat i +y\ \widehat j +z\ \widehat k [/tex]

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

[tex] \nabla \times \vec v [/tex]

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

1. What does the equation "Curl v = 2w_k" mean?

The equation "Curl v = 2w_k" is a vector equation in three-dimensional space that represents a mathematical operation known as the curl. The symbol "v" represents a vector quantity, and "w_k" represents a vector field in the k direction. The number 2 in front of the vector field indicates the strength of the field.

2. How do I interpret the result of Curl v = 2w_k?

The result of Curl v = 2w_k is another vector quantity that represents the direction and strength of the curl operation on the vector field w_k. The direction of the resulting vector is perpendicular to the plane formed by the direction of the vector field and the direction of the curl operation. The length of the resulting vector represents the strength of the curl.

3. What is the significance of the curl operation in physics?

The curl operation is an important concept in physics, specifically in the field of electromagnetism. It is used to describe the circulation or rotation of a vector field in three-dimensional space, which is relevant in many physical phenomena such as fluid flow, magnetic fields, and electric fields.

4. Can you provide an example of a physical application of the equation Curl v = 2w_k?

Sure, an example of a physical application of Curl v = 2w_k is in the study of fluid mechanics. The curl of a velocity field can be used to determine the vorticity, which represents the local spinning of a fluid element. This is useful in understanding the behavior of fluids, such as in weather patterns or in the motion of ocean currents.

5. How do I solve problems involving Curl v = 2w_k?

In order to solve problems involving Curl v = 2w_k, you will need to have a good understanding of vector calculus and how to perform the curl operation. It is also helpful to have a solid understanding of the physical application of the equation, as well as a strong grasp of mathematical techniques for solving vector equations. It is recommended to practice with various examples and seek guidance from a teacher or textbook if needed.

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