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Deriving the formula: v=wr

  1. Oct 15, 2014 #1
    How to derive the formula:
    v=wr
    where v is the tangential velocity, w is the rotational velocity, and r i the radius vector?

    From the attached image, it can be concluded that (each quantity is a vector): w=r x v, also v=w x r, and r= v x w. All three vectors are perpendicular to each other, therefore the intensity of each vector can be calculated by vector multiplication. Then (each quantity is a vector modulus):
    w=rv, v=wr, r=vw, this system of equations is true if w=v=r which mustn't be true. I need an explanation. What did I wrong to arrive at this incorrect equality?
     

    Attached Files:

  2. jcsd
  3. Oct 15, 2014 #2

    ShayanJ

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    Look at the image below:
    circle_arc.png

    By definition of a radian(unit of angle), we can write [itex] s= r \theta [/itex](where [itex] \theta [/itex] is in radians). Now, assuming a fixed radius, differentiation of the equation w.r.t. time will give you the desired result.
     
  4. Oct 15, 2014 #3

    SteamKing

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    The angular velocity ω is usually a scalar, rather than a vector quantity. The magnitude of the radial velocity is given as v = ω r, where r is the magnitude of the radius vector.

    For a derivation of the radial velocity vector, see this article:

    http://en.wikipedia.org/wiki/Circular_motion

    and note the difference between ω and the vector Ω.

    You should also be aware that the cross product does not commute, so that all of these statements may not be valid simultaneously:

    w=r x v, v=w x r, r= v x w.
     
  5. Oct 15, 2014 #4
    Okay, but I want to derive it the way I previously posted, but it brings me nowhere and I want to make myself clear what was wrong.
    The cross product does not commute, but how does that explain anything?
     
  6. Oct 15, 2014 #5

    A.T.

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    That looks more like tangential velocity to me. Radial velocity is along the radius:
    http://en.wikipedia.org/wiki/Radial_velocity
     
  7. Oct 15, 2014 #6

    A.T.

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    How did you conclude that from the image? It's only true if they are all unit vectors
     
  8. Oct 15, 2014 #7
    I think that you have the answer, but please explain it. How is it true only then?
     
  9. Oct 15, 2014 #8

    A.T.

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    You have to explain how you concluded all that from your picture.
     
  10. Oct 15, 2014 #9
    I wrote it. w=r x v, also v=w x r, and r= v x w from the picture. For example r x v gives the vector w. I got all by calculating the vector product:
    <r x v>=<w>=rv*sin(pi/2)=rv. The same way I got that v=wr, r=vw.
     
  11. Oct 15, 2014 #10

    DrGreg

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    Your picture only shows three vectors perpendicular to each other. It doesn't tell you what the lengths of the vectors are. The correct equations should be[tex]
    \textbf{v} = \boldsymbol{\omega} \times \textbf{r}; \;
    \boldsymbol{\omega} = \frac{\textbf{r} \times \textbf{v}}{r^2}
    [/tex]
     
  12. Oct 16, 2014 #11
    How did you get that w=rxv/r^2?
     
  13. Oct 16, 2014 #12

    A.T.

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    None of this follows from the picture. Just because 3 vectors are perpendicular, doesn't mean they necessarily represent the operands and result of a vector product.
     
  14. Oct 16, 2014 #13

    DrGreg

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    See, for example, angular velocity, or any text book on the subject.
     
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