Relationship between linear velocity and angular velocity

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SUMMARY

The relationship between linear velocity and angular velocity is defined by the equation v = ωr, where v is the linear velocity, ω is the angular velocity, and r is the radius. The discussion clarifies that only the tangential component of velocity, which is perpendicular to the radius, is relevant for this equation. The confusion arises from the inclusion of the cosine factor, which is unnecessary when considering the limit as θ approaches 0. The correct interpretation emphasizes that the radial component does not affect the angular motion.

PREREQUISITES
  • Understanding of angular velocity (ω) and linear velocity (v)
  • Familiarity with trigonometric functions, particularly cosine (cos)
  • Basic knowledge of circular motion and radius (r)
  • Ability to analyze limits in calculus
NEXT STEPS
  • Study the derivation of the equation v = ωr in the context of circular motion
  • Learn about the role of tangential and radial components in motion analysis
  • Explore the implications of limits in trigonometric functions, especially as they relate to angular motion
  • Investigate common misconceptions in physics regarding velocity components
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Students of physics, educators teaching mechanics, and anyone interested in understanding the principles of angular and linear velocity relationships.

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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