Understanding Complex Vector Rotation

[jeff1evesque]

Homework Statement

Can someone explain to me why complex vectors of the following form will rotate counterclockwise in the x-y plane [with a velocity of w],

\hat{v}(t) = cos(\omega t)\hat{x} + sin(\omega t)\hat{y} (#1)

And why the following equation, the unit vector rotates in the clockwise direction in the x-z plane,

\hat{v}(t) = -sin(\omega t)\hat{x} + cos(\omega t)\hat{z} (#2)2. An attempt:
So I approached this problem by assuming \omega = 1; t = 0, t = \frac{\pi}{2}, t= \pi, t = \frac{3\pi}{2}, t = 2\pi. When I computed this values, it seemed to me \hat{v}(t) increased in the counterclockwise direction for both equations (#1) and (#2). Can someone explain to me the nature of the rotation for complex vectors?

thanks,JL

 

[D H]

Where are the complex numbers in this problem? It appears you are talking about rotation in real 3-space.

What do you mean by clockwise and counterclockwise? Imagine a transparent clock, one whose hands are visible from behind. When viewed from the back the hands will appear to be moving counterclockwise.

The right hand rotation rule provides a much better basis for terminology. In the first problem, the angular velocity vector points along +z axis. Where does it point in the second problem?

 

[jeff1evesque]

Where are the complex numbers in this problem? It appears you are talking about rotation in real 3-space.

What do you mean by clockwise and counterclockwise? Imagine a transparent clock, one whose hands are visible from behind. When viewed from the back the hands will appear to be moving counterclockwise.

The right hand rotation rule provides a much better basis for terminology. In the first problem, the angular velocity vector points along +z axis. Where does it point in the second problem?

I think these "Complex Vectors" is in relation to "Phasors". When i say the orientation, for example clockwise, I mean the unit vector \hat{v}(t) is rotating about a chosen plane [for instance x-y plane] in that respective direction. As it rotates, [since it has a fixed magnitude], it will trace a circle as the angular velocty \omega increases. I hope that helps.

 

[D H]

You don't need to help me. You are the one asking the question. This looks like homework, so I cannot by the rules of this site tell you the answer straight out. Besides, the question is ill-formed. What do you mean by clockwise? Think of the transparent clock.

 

[jeff1evesque]

You don't need to help me. You are the one asking the question. This looks like homework, so I cannot by the rules of this site tell you the answer straight out. Besides, the question is ill-formed. What do you mean by clockwise? Think of the transparent clock.

If I cannot help you understand my question, how will I get my question answered? I am simply typing up definitions from my notes and asking for an interpretation from a different source- and right now that is you. And to my knowledge, the complex vector will rotate in the clockwise direction, by the equation (#1) of above. I am trying to make sense of my notes, and I am having difficulty right now, sorry for conjuring up bad questions, as I've never taken any EE courses before but am interested.

Thanks,JL

 

[D H]

It isn't a complex vector. You have no complex numbers.

At t=0, the vector in the first equation will be oriented along the plus-x axis. A short time later, the vector's x component won't have changed much and the y component will be small but positive. The reason this first equation is said to rotate clockwise is because the x and y axes are typically represented as being horizontal and vertical, respectively, and with increasingly positive values to the right and upward. If you drew the +x axis to the left and the +y axis down, you would get the opposite picture.

The second problem: What made you use the x-z plane, as opposed to the z-x plane? What direction are the axes oriented?

 

[Redbelly98]

(1) is reasonably unambiguous, because there is a preferred convention for viewing the x-y plane (+x is to the right, +y is upward). You can think of this as being viewed from the positive z side of the x-y plane.

However, for (2), it depends on what direction the x-z plane is to be viewed from: positive y or negative y side of the plane?

Is there a figure with the problem statement that would clarify this?

 

[jeff1evesque]

It isn't a complex vector. You have no complex numbers.

At t=0, the vector in the first equation will be oriented along the plus-x axis. A short time later, the vector's x component won't have changed much and the y component will be small but positive. The reason this first equation is said to rotate clockwise is because the x and y axes are typically represented as being horizontal and vertical, respectively, and with increasingly positive values to the right and upward. If you drew the +x axis to the left and the +y axis down, you would get the opposite picture.

The second problem: What made you use the x-z plane, as opposed to the z-x plane? What direction are the axes oriented?

It says "Write the equation of a unit vector that rotates clockwise in the x-z plane when viewed from the positive y-axis. The vector should poin in the z direction at t = 0. Then couple lines below, it tells me,
"In order that it rotates clockwise: <br /> \hat{v}(t) = -sin(\omega t)\hat{x} + cos(\omega t)\hat{z}<br />."

 

[D H]

There you go. Right hand rotation rule, which you can google. If you don't understand what you find, ask away.

 

[jeff1evesque]

There you go. Right hand rotation rule, which you can google. If you don't understand what you find, ask away.

I think I knew the right hand rule- or hope so, but I am trying to think of why the equation itself causes the vector to rotate. So I let wt = 0, and thus z = 1, when I let wt = pi/2 then x = -1. If I continue on, it seems to me as values inside the equation increases, a rotation is clockwise. Is that a reasonable way of thinking? And can I apply this way of thinking to the other equation that I asked about?

thanks,


Jeff

 

[Redbelly98]

Yes, that's a good way to look at it. ωt = 0, π/2, π, and 3π/2. It should be clear by the time you get to π or even π/2.

 

[jeff1evesque]

Yes, that's a good way to look at it. ωt = 0, π/2, π, and 3π/2. It should be clear by the time you get to π or even π/2.

Oh cool, thanks a lot. I actually was thinking in this way earlier, but as I was doing it, I was picturing the x-axis as the opposite orientation as it should be, and that just threw me in a loop haha. Thanks for the help guys.