Deriving maximum velocity in a rotated frame

Click For Summary

Discussion Overview

The discussion revolves around deriving the maximum velocity of a particle in a rotated uv coordinate frame, given its maximum velocities along the x and y axes in the Cartesian plane. The focus is on understanding how to translate these velocities into the new frame.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant describes a particle's maximum velocities along the x and y axes as A and B units, respectively, and seeks guidance on deriving the corresponding maximum velocities in the rotated uv plane.
  • Another participant suggests using a diagram to represent the vector from the origin to the point (A, B) and drawing perpendiculars to the u and v axes to find the coordinates in the new frame.
  • A subsequent reply clarifies the need to draw a line representing the maximum velocity vector and emphasizes the importance of identifying the coordinates of this vector in the (u, v) system, noting that one of the coordinates may be negative.
  • The original poster expresses understanding of the suggestions provided.

Areas of Agreement / Disagreement

Participants appear to agree on the approach of using a diagram to visualize the problem, but there is no explicit consensus on the final method of calculation or the implications of the negative coordinate.

Contextual Notes

The discussion does not resolve the mathematical steps required to derive the maximum velocities in the rotated frame, and assumptions about the angle between the axes are not explicitly stated.

Who May Find This Useful

Individuals interested in vector transformations, coordinate systems, and applications in physics or engineering may find this discussion relevant.

jumbo1985
Messages
19
Reaction score
1
rotate%20frame_zps6lbjzvxf.jpg

I have a particle that travels in the cartesian plane with the maximum velocity of A units along the x-axis and B units along the y-axis per unit of time.

How do I go about deriving the maximum velocity of my particle in the rotated uv plane? (the maximum distance the particle can along the u and the v axes in one unit of time)

Any tips greatly appreciated.
 
Physics news on Phys.org
Hi jumbo:

I suggest it might help if you add an arrow in the diagram to represent the vector M from the origin to point (x=A,y=B). Then draw perpendiculars from (A,B) to the axes u and v. If you know the angle between x-axis and u axis, you can calculate the length of the perpendiculars to the u axis and v axis which represent the M coordinates with respect to the v axis and u axis respectively.

Hope this helps.

Regards,
Buzz
 
Thanks Buzz Bloom.

Is this what you mean?

vel_max_zpsuvcup2vg.jpg


Where the length of A' is the maximum distance my particle can travel along the u-axis of the rotated frame in one unit of time. Similarly, the length of B'...
 
Last edited:
Hi jumbo:

I apologize for not expressing my thought more clearly.

You want to draw a line, call it L, from the origin to the point M. M will be on the vertical green line at the height B. L represents the vector corresponding to the maximum velocity of the particle you are describing.

Next draw lines from the point M perpendicular to the u and v axes. These lines will show the coordinates of the vector L in the (u,v) coordinate system. The resulting diagram should help you figure out how to calculate the u and v coordinates of the vector L.

By the way, for the particular example you have drawn, you may want to notice that the v coordinate of L is negative.

Hope this helps.

Regards,
Buzz
 
I think I got it. Thanks Buzz

vel_max_2_zpscfsvfczk.jpg
 
  • Like
Likes   Reactions: Buzz Bloom

Similar threads

  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
979
  • · Replies 23 ·
Replies
23
Views
2K