Finding the curl of velocity in spherical coordinates

AI Thread Summary
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
Messages
37
Reaction score
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

  • methods 3.jpg
    methods 3.jpg
    26.1 KB · Views: 685
Last edited:
Physics news on Phys.org
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top