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Uniform circular motion?

  1. Jul 6, 2016 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    science2.PNG
    How is Δθ in circle equals to angle in velocity vectors triangle?
    I tried using simple geometry but I can't.
     
  2. jcsd
  3. Jul 6, 2016 #2

    Doc Al

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    Staff: Mentor

    Try this:

    What if r' were right on top of r? What is the angle between them? What is the angle between the velocity vectors?

    Now move r' just a bit, an angle Δθ from r. How does the angle between the velocity vectors change?
     
  4. Jul 6, 2016 #3
  5. Jul 6, 2016 #4
  6. Jul 6, 2016 #5
    Its because they are angles that have sides vertical to each other. r' is vertical to v' and r is vertical to v so the angle of (r',r) equals the angle of (v',v).
     
  7. Jul 6, 2016 #6
    Delta2 and Clara I appreciate your explaination but can you please describe some '90 degree usage'.
     
  8. Jul 6, 2016 #7

    Doc Al

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    Another way to look at it: Realize that the velocity vectors are always 90° rotated compared to the r vectors. So if the angle that an r vector makes with the x-axis is θ, then the corresponding velocity vector must make an angle of θ + 90°.
     
  9. Jul 6, 2016 #8
    In the photo, θ+a = 90..(1) , a+b=90...(2), (1)-(2), θ=b , so it proves your statement "Δθ in circle equals to angle in velocity vectors triangle".
    I hope it helps.
     
  10. Jul 8, 2016 #9

    James R

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    One other point: the vector diagram on the right does not correspond to the vectors shown on the left.
    Recall that ##\Delta \vec{v} = \vec{v}'-\vec{v}##. You also seem to have swapped the ##\vec{v}'## and the ##\vec{v}## in the diagram.
     
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