# Passive Rotation

1. Sep 30, 2008

### Soff

When you rotate passivly an arbitrary vector in $$P$$ (which is not in the origin of a coordinate system), you will get the following new coordinates for the same vector:

$$x'=x \cos\theta+y\sin \theta$$
$$y'=\frac{1}{r}(y\cos \theta-x\sin\theta)$$

where $$r$$ is the distance from the origin to the point $$P$$ and $$x,y$$ are the components of the vector. Can somebody explain me, why you have to divide by $$r$$ in the second equation?

2. Sep 30, 2008

### HallsofIvy

Staff Emeritus
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:
$$x'= x cos(\theta)+ y sin(\theta)$$
$$y'= y cos(\theta)- x sin(\theta)$$
which are simply what you give without the "r".

You can check by taking x= 2, y= 0 and rotating through $\theta= 90$ 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.

3. Oct 2, 2008

### Soff

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 $$A^{\theta}$$ and $$A^{r}$$. The professor told me that the solution is:

$$A^{r}=A^{x}\cos\theta+A^{y}\sin\theta$$
$$A^{\theta}=\frac{1}{r}(A^{y}\cos\theta-A^{x}\sin\theta)$$

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

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4. Oct 2, 2008

### Doodle Bob

Why don't you ask the professor?