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Passive Rotation

  1. Sep 30, 2008 #1
    When you rotate passivly an arbitrary vector in [tex]P[/tex] (which is not in the origin of a coordinate system), you will get the following new coordinates for the same vector:

    [tex]x'=x \cos\theta+y\sin \theta[/tex]
    [tex]y'=\frac{1}{r}(y\cos \theta-x\sin\theta)[/tex]

    where [tex]r[/tex] is the distance from the origin to the point [tex]P[/tex] and [tex]x,y[/tex] are the components of the vector. Can somebody explain me, why you have to divide by [tex]r [/tex] in the second equation?
     
  2. jcsd
  3. Sep 30, 2008 #2

    HallsofIvy

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    You don't! For one thing, there is no "r" in the original information- unless you are assuming that r is the length of the vector <x, y>. But, in any case, there is no "r" in the correct formulas:
    [tex]x'= x cos(\theta)+ y sin(\theta)[/tex]
    [tex]y'= y cos(\theta)- x sin(\theta)[/tex]
    which are simply what you give without the "r".

    You can check by taking x= 2, y= 0 and rotating through [itex]\theta= 90[/itex] degrees.
    Assuming, again, that r is the length of the vector, then here r= 2 and your formulas give x'= 2(0)+ 0(1)= 0, y'= (1/2)(2(1)- 1(0))= 1 but <0, 1> is a a vector with length 1.

    Using the formulas without the r, you get x'= 2(0)+ 0(1)= 0, y'= 2(1)- 1(0)= 2 giving the vector <0, 2> which is correct.
     
  4. Oct 2, 2008 #3
    I attached the graphic which describs the situation with the rotation. What I want is to describe the vector in the lower part of the system in coordinates of [tex]A^{\theta}[/tex] and [tex]A^{r}[/tex]. The professor told me that the solution is:

    [tex]A^{r}=A^{x}\cos\theta+A^{y}\sin\theta[/tex]
    [tex]A^{\theta}=\frac{1}{r}(A^{y}\cos\theta-A^{x}\sin\theta)[/tex]

    The first component is simply a rotation, but why this factor [tex]\frac{1}{r}[/tex] in the second component?
     

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  5. Oct 2, 2008 #4
    Why don't you ask the professor?
     
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