Understanding Polar Coordinate Unit Vectors

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Homework Help Overview

The discussion revolves around understanding polar coordinate unit vectors, specifically the roles of the unit vectors \( \hat{e}_r \) and \( \hat{e}_{\theta} \) in expressing position and velocity in polar coordinates.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the meaning of expressing position in terms of polar unit vectors and question the nature of \( \hat{e}_{\theta} \) as it relates to direction. There are discussions on how polar coordinates differ from Cartesian coordinates, particularly regarding the constancy of unit vectors and the implications for position and velocity.

Discussion Status

Several participants have provided insights into the nature of polar coordinates and the relationships between the unit vectors. There is an ongoing exploration of how to express velocity in terms of \( \hat{e}_r \) and \( \hat{e}_{\theta} \), with some guidance on the derivatives of these unit vectors being discussed. Multiple interpretations of the problem are being examined, particularly regarding the application of derivatives in polar coordinates.

Contextual Notes

Participants note the need to consider the derivatives of the unit vectors when differentiating position in polar coordinates. There is also mention of specific values and equations from dynamics classes that may influence the discussion.

  • #31
Yes, but does it move radially outward now?

ehild
 
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  • #32
I am not grasping what you are trying to ask...
 
  • #33
The particle moves along a circle of radius R=5. You want its velocity vector in terms of Er and Eθ. Er is multiplied by dr/dt. r is the distance of the particle from the centre. Does that distance change while the particle moves along a circle? So what is the value of dr/dt?

ehild
 
  • #34
it is 0 because dr/dt of a constant = 0
 
  • #35
dr/dt is not 5. It is zero, as the distance from the origin does not change. So how can you write the velocity vector? And what about the acceleration?

ehild
 
  • #36
Yea after I wrote the wrong answer, I thought about it and changed it. But the velocity would just be this right? v = 5*(dθ/dt)*Eθ and the acceleration would be 0?
 
  • #37
Yes, the velocity vector is \vec v = 10 \hat e _{\theta} in this case.

The acceleration is the time derivative of the velocity. It is not zero, as the velocity changes direction. You need the derivative of Eθ now. Go back to #20 and figure out how is it related to Er.

ehild
 
  • #38
I only needed the velocity vector in polar coordinate form. Thank you so much for the help!
 
Last edited:

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