Finding the curl of velocity in spherical coordinates

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The discussion revolves around evaluating the linear velocity vector and its curl in both cylindrical and spherical coordinates for a rigid object rotating about the z-axis. Participants clarify the conversion of the angular velocity vector ω into spherical coordinates, noting that since ω lies along the z-axis, it can be expressed as ω = ω r-hat. There is confusion regarding the cross product of ω and the position vector R in spherical coordinates, leading to discussions about the correct representation of unit vectors. Suggestions are made to substitute z-hat with its spherical components to facilitate the calculations. The conversation also includes a request for resources on using LaTeX for formatting equations.
John004
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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.
 

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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
 
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?
 
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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