Finding the curl of velocity in spherical coordinates

In summary: You can also write it as ##\vec{\omega} = \omega_{z}\hat{k} = \omega\hat{k}##, where ##\hat{k}## is the unit vector along the z-axis. Then, at an arbitrary point in space, you can write ##\vec{\omega} = \omega \hat{k} = \omega (\hat{r}\cos\theta - \hat{\theta}\sin\theta)##. This will give you a non-zero value for v in spherical coordinates. In summary, the conversation discusses the angular and linear velocity vectors of a rigid object rotating about the z-axis. It also covers the evaluation of these vectors in cylindrical and spherical coordinates, as well as the
  • #1
John004
37
0

Homework Statement


The angular velocity vector of a rigid object rotating about the z-axis is given by
ω = ω z-hat. At any point in the rotating object, the linear velocity vector is given by v = ω X r, where r is the position vector to that point.

a.) Assuming that ω is constant, evaluate v and X v in cylindrical coordinates.

b.) Evaluate v in spherical coordinates.

c.) Evaluate the curl of v in spherical coordinates and show that the resulting expression is equivalent to that given for X v in part a.

Homework Equations


The expressions for the curl in cylindrical and spherical coordinates. Since I don't know how to put the determinant here ill just leave them out.

For spherical

x = r sinθ cosΦ

y = r sinθ sinΦ

z = r cos θ

The Attempt at a Solution


So I worked out part a correctly (I think) which is in the attachment, but I'm stuck on part b.

b.) So for this part I have to convert ω to spherical coordinates. Since ω only lies along the z-axis, that means that Φ and θ are equal to zero, so

ω = ω r-hat

and the position vector in spherical polars is

R =
r r-hat

so that means that when I cross ω and R I get zero, I don't know what I'm missing.
 

Attachments

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  • #2
part (a) looks good.
John004 said:
b.) So for this part I have to convert ω to spherical coordinates. Since ω only lies along the z-axis, that means that Φ and θ are equal to zero, so

ω = ω r-hat

Go to an arbitrary point in space and try to write ω in terms of the unit vectors ##\hat{r}##, ##\hat{\theta}##, and ##\hat{\phi}## at that point.

and the position vector in spherical polars is

R =
r r-hat
OK
 
  • #3
TSny said:
part (a) looks good.Go to an arbitrary point in space and try to write ω in terms of the unit vectors ##\hat{r}##, ##\hat{\theta}##, and ##\hat{\phi}## at that point.

OK
Is there some resource you can point me to so I can learn how to type out symbols and equations neatly like you just did?

I can't really picture it in the way you're asking me too. What if I substitute z-hat = r-hat cosθ - θ-hat sinθ?
 
  • #4
John004 said:
Is there some resource you can point me to so I can learn how to type out symbols and equations neatly like you just did?
https://www.physicsforums.com/help/latexhelp/
You can learn a lot by just examining how others have used Latex in their posts. That's how I picked it up. I still have a lot to learn.

I can't really picture it in the way you're asking me too. What if I substitute z-hat = r-hat cosθ - θ-hat sinθ?
Yes, that's it.
 

What is the curl of velocity in spherical coordinates?

The curl of velocity in spherical coordinates is a measure of the rotation or swirl in a fluid flow at a particular point. It is a vector quantity that describes the direction and magnitude of the rotation at that point.

How do you calculate the curl of velocity in spherical coordinates?

The curl of velocity in spherical coordinates can be calculated using the following formula:
curl(V) = (1/r*sin(theta)) * (d/dtheta(sin(theta)*V_phi) - d/dphi(V_theta)) +
(1/r) * (d/dphi(V_r*sin(theta)) - d/dr(V_phi)) +
(1/r) * (V_r * d/dphi(sin(theta)))
where V_r, V_theta, and V_phi are the components of the velocity vector in the radial, polar, and azimuthal directions, respectively.

What does the curl of velocity tell us about fluid flow in spherical coordinates?

The curl of velocity provides information about the rotation or swirling motion in a fluid flow at a particular point in spherical coordinates. It can help us understand the direction and magnitude of the rotation, as well as any potential vortices or eddies present in the flow.

Why is it important to find the curl of velocity in spherical coordinates?

Finding the curl of velocity in spherical coordinates is important in fluid dynamics and other areas of physics, as it helps us understand the complex motion of fluids in three-dimensional space. It can also be used to analyze and predict the behavior of fluids in various scenarios, such as in weather patterns or in engineering designs.

What are some real-world applications of finding the curl of velocity in spherical coordinates?

The curl of velocity in spherical coordinates has many practical applications, such as in meteorology for predicting weather patterns, in oceanography for studying ocean currents, and in engineering for designing efficient and safe fluid systems. It is also used in physics research to study the behavior of fluids in various contexts, such as in astrophysics or geophysics.

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