Relationship between linear velocity and angular velocity

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Homework Help Overview

The discussion revolves around the relationship between linear velocity and angular velocity, specifically addressing the equation v = wr and its variations. Participants are examining the conditions under which these equations hold true and the implications of angular displacement on linear velocity.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants explore the validity of the equation v = wr versus v = wr*cos(theta), questioning the role of the angle theta in determining the correct relationship. Some participants analyze the components of velocity and their relevance to the angular motion.

Discussion Status

The discussion is active, with participants providing insights and questioning each other's reasoning. There is a focus on clarifying misunderstandings regarding the equations and the conditions under which they apply. Some participants express differing views on the correctness of the provided answers, indicating a lack of consensus.

Contextual Notes

Some participants suggest that the original problem statement may not have been fully conveyed, which could affect the interpretation of the equations discussed. The limits of theta are also being examined to understand the behavior of the equations in specific cases.

good jelly
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Homework Statement
Find the relationship between the linear velocity and angular velocity for mass B which is moving about mass A (mass A is stationary).
Relevant Equations
v = wr
I find the velocity vector that is perpendicular to the radius which is v*cos(theta). Once I obtained the velocity that is perpendicular to the radius I used the equation : v = wr. But my answer seems to be wrong, the actual answer to this question is v = wr*cos(theta). Why?

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good jelly said:
the actual answer to this question is v = wr*cos(theta)
No it isn’t and it is easy to see that it isn’t by considering the limit ##\theta \to 0##.
 
good jelly said:
Why?
Only the velocity component perpendicular to the radius (i.e. tangential to the path) counts. The radial component (towards or away from the center) does not count because it does not change the angle.
 
@good jelly :
Welcome to PF!

Have you given the complete statement of the problem "word-for-word"?
Based on your diagram, your equation ##v_B \cos \theta = \omega R## looks correct.
 
kuruman said:
Only the velocity component perpendicular to the radius (i.e. tangential to the path) counts. The radial component (towards or away from the center) does not count because it does not change the angle.
That wasn’t the question. That was exactly what OP did. The question was why the wrong answer provided by [unknown, probably teacher] is correct. It isn’t.
 
Orodruin said:
That wasn’t the question. That was exactly what OP did. The question was why the wrong answer provided by [unknown, probably teacher] is correct. It isn’t.
You are right, that was not the question. One can only provide reassurance that OP's answer is correct without attempting to divine like a haruspex why the other answer is incorrect.
 
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Orodruin said:
No it isn’t and it is easy to see that it isn’t by considering the limit ##\theta \to 0##.
Isn't that limit the one case in which the two formulae both yield ##v=\omega r##?
Maybe you meant ##\theta\rightarrow \pi/2##.
 
haruspex said:
Isn't that limit the one case in which the two formulae both yield ##v=\omega r##?
Maybe you meant ##\theta\rightarrow \pi/2##.
Yes, definitely. Good catch.
 

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