Question about translating frame

[issacnewton]

Hello

I have a question about the meaning of this term "translating frame". Let's consider x-y axis fixed in one place. Now we can construct a unit circle with origin at (0,0). Now I am going to construct another frame of reference with origin at (1,0) (in x-y system). I will let \mathbf{x}^{\prime} axis be parallel to \mathbf{x} axis and \mathbf{y}^{\prime} axis be parallel to
\mathbf{y} axis. Now I will let this new frame's origin go around unit circle with no change in the direction of \mathbf{x}^{\prime} and \mathbf{y}^{\prime} axis. So will this be a translating frame or rotating frame with respect to the fixed frame. I think its a translating frame, since the axes are not changing directions. Am I right ?

thanks

 

[ash64449]

If i consider the meaning of translation and rotation to be same that we consider in dealing with the motion of the bodies,

The situation is similar to a body revolving around a point.

Do you consider this situation translation or rotation?

 

[issacnewton]

ash64449, body revolving around a point would be rotation. But the way I am considering it, the direction of axes inside the body is not changing, so I would say that body's frame is translating with respect to the fixed frame

 

[ash64449]

he direction of axes inside the body is not changing,

Yes. the direction of the axes is not changing. But i thought a body is said to be rotating if every point of the body revolves in a circle whose center lies in the axis of rotation.

That is here the priority is given to the point of the body. In this case we can see that every point of the new reference frame revolves around the origin of the old frame.

I may be wrong. I think you must wait for some other's opinion. if the analogy i consider is correct, i may be correct.

 

[jbriggs444]

Now I will let this new frame's origin go around unit circle with no change in the direction of \mathbf{x}^{\prime} and \mathbf{y}^{\prime} axis. So will this be a translating frame or rotating frame with respect to the fixed frame. I think its a translating frame, since the axes are not changing directions.

"Translating" and "rotating" are just words we sometimes use to briefly describe two kinds of motion of one frame of reference with respect to another. Those two kinds of motion are not the only possibilities. I am not aware of a term to describe the sort of circular motion that you have in mind. One possible word would be "oscillating".

 

[BobG]

Hello

I have a question about the meaning of this term "translating frame". Let's consider x-y axis fixed in one place. Now we can construct a unit circle with origin at (0,0). Now I am going to construct another frame of reference with origin at (1,0) (in x-y system). I will let \mathbf{x}^{\prime} axis be parallel to \mathbf{x} axis and \mathbf{y}^{\prime} axis be parallel to
\mathbf{y} axis. Now I will let this new frame's origin go around unit circle with no change in the direction of \mathbf{x}^{\prime} and \mathbf{y}^{\prime} axis. So will this be a translating frame or rotating frame with respect to the fixed frame. I think its a translating frame, since the axes are not changing directions. Am I right ?

thanks

It's a translating frame.

Similar to an inertial geocentric-equatorial reference frame moving around the Sun.

(You didn't provide much context for your question, and I'm not sure everyone understood your question.)

 

[issacnewton]

Hello Bob, I think the situation I explained is simple enough. One frame with x-y axes is fixed. And another frame with primed x-y axes is revolving around it, but the old and new axes remain parallel. So this should be considered translating frame. Since the unit vectors are not changing direction, we won't have to differentiate them when we take higher derivatives for velocity and acceleration...

 

[Dale]

I probably wouldn't call it translating or rotating or anything else. I would just write down the transformation equations. Anything else is bound to be misunderstood. English is just not sufficiently precise to communicate this clearly in a small number of words.

 

[A.T.]

Hello

I have a question about the meaning of this term "translating frame". Let's consider x-y axis fixed in one place. Now we can construct a unit circle with origin at (0,0). Now I am going to construct another frame of reference with origin at (1,0) (in x-y system). I will let \mathbf{x}^{\prime} axis be parallel to \mathbf{x} axis and \mathbf{y}^{\prime} axis be parallel to
\mathbf{y} axis. Now I will let this new frame's origin go around unit circle with no change in the direction of \mathbf{x}^{\prime} and \mathbf{y}^{\prime} axis. So will this be a translating frame or rotating frame with respect to the fixed frame. I think its a translating frame, since the axes are not changing directions. Am I right ?

thanks

The second frame not a rotating frame, so it doesn't have centrifugal or Coriolis forces.

But it's also not an inertial frame (if the first frame one was inertial) because it undergoes centripetal acceleration. So it does have a uniform time-dependent inertial force field.

This is discussed here:
http://www.vialattea.net/maree/eng/index.htm

Rotating frame with an inertial origin has a radial centrifugal force field:

Non-rotating frame with non-inertial origin (your case) has a uniform time-dependent force field:

Rotating frame with an non-inertial origin has both:

 

[PWiz]

Do you mean that your new frame moves around the origin (along the circle) without changing orientation, as if it were sitting on a passenger seat on a ferris wheel with the origin at the center of the wheel?

If yes, then this is a translation - the "length" of the translation vector (from the origin of the unprimed frame of course) stays invariant, but the angle that the vector subtends from the x/y axis will be a function of time that varies (linearly if the rotation has a constant speed) because the x and y components of the vector are changing with time. Just construct a vector equation and see for yourself.

 

[issacnewton]

I think A.T. has described what I am saying in the second animation. I will read the link posted.