1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

    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:


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

    Attached Files:

  5. Oct 2, 2008 #4
    Why don't you ask the professor?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook