# Rotational Frame Transformations

• LAP3141
In summary, the conversation discusses the transformation of velocity components in a rotating reference frame. The method described assumes that the velocity components are the same in both frames, but this is not the case. A more advanced approach is needed, such as transforming the velocity components into the stationary frame and then back into the rotating frame. The conversation also includes links to related web pages and a PDF file for further understanding.
LAP3141
Consider a rotating disk with the center at the origin of a stationary Cartesian coordinate system, (x, y).

At t = 0, on the circumference of the disk, someone/something shoots a particle with constant velocity components Dvx, Dvy (where the D indicates the rotating disk). Also at time t=0, the disk coordinate of the shooter lines up with the Cartesian (R, 0) coordinate.

Thus, the initial points are the same for both frames:
(x0, y0) = (Dx0, Dy0) = (R, 0).

Also the initial, and constant, velocity components are equal in both frames:
[vx0, vy0] = [Dvx0, Dxy0].

In the Cartesian frame, the particle trajectory is a straight line:
x = x0 + vx0 * t
y = y0 + vy0 * t

However, if I transform this Cartesian (or inertial) trajectory into the rotating frame using the matrix:

| cos(w*t) sin(w*t) |
| -sin(w*t) cos(w*t) |

where w is the angular rotation velocity, I do not get the expected trajectory as seen in the rotating frame. I get a curved trajectory but the trajectory is not what is should be.

The initial velocity has to also be transformed, but since the initial velocity is the same in both frames, and since the velocity is constant, I don't understand intuitively why this is so.

LAP3141 said:
Summary: Why does a constant velocity have to be transformed when dealing with rotating reference frames?

Consider a rotating disk with the center at the origin of a stationary Cartesian coordinate system, (x, y).

At t = 0, on the circumference of the disk, someone/something shoots a particle with constant velocity components Dvx, Dvy (where the D indicates the rotating disk). Also at time t=0, the disk coordinate of the shooter lines up with the Cartesian (R, 0) coordinate.

Thus, the initial points are the same for both frames:
(x0, y0) = (Dx0, Dy0) = (R, 0).

Also the initial, and constant, velocity components are equal in both frames:
[vx0, vy0] = [Dvx0, Dxy0].

In the Cartesian frame, the particle trajectory is a straight line:
x = x0 + vx0 * t
y = y0 + vy0 * t

However, if I transform this Cartesian (or inertial) trajectory into the rotating frame using the matrix:

| cos(w*t) sin(w*t) |
| -sin(w*t) cos(w*t) |

where w is the angular rotation velocity, I do not get the expected trajectory as seen in the rotating frame. I get a curved trajectory but the trajectory is not what is should be.

The initial velocity has to also be transformed, but since the initial velocity is the same in both frames, and since the velocity is constant, I don't understand intuitively why this is so.
What do you mean by not what it should be? Is it upside down?

LAP3141 said:
Also the initial, and constant, velocity components are equal in both frames:
This is not correct. The tangential velocity is subtracted in the rotating frame.
LAP3141 said:
I get a curved trajectory but the trajectory is not what is should be.
The method you describe is valid, so either you made an algebra mistake or your expectation of what it should be is wrong.

Dale said:
The method you describe is valid, so either you made an algebra mistake or your expectation of what it should be is wrong.

I reckon it will be upside down.

Dale said:
This is not correct. The tangential velocity is subtracted in the rotating frame.

The method you describe is valid, so either you made an algebra mistake or your expectation of what it should be is wrong.

OK. The velocity components are not the same in both frames.

After finding some more advanced descriptions on the web I am beginning to see the reason more clearly.

But the method I described is also NOT correct because it assumes that the velocity components are the same.

I started all this after discovering these related web pages:

http://www.phermi.com/space-station-catch
http://www.phermi.com/space-station-catch-solution

In the solution given, the velocity components in the rotating frame are transformed into the stationary frame and these transformed components are used to determine the trajectory in the stationary frame. Then the stationary frame trajectory is transformed back into the rotating frame.

This procedure is a bit awkward and I am trying to learn a more systematic approach.

But the solution which I described does NOT produce the curves shown in the solution web page. This is due to the assumption that the velocity components are equal in both frames.

I will study some more advanced web pages until I get the correct mathematics.

Unfortunately, the web pages which I cited do not properly load MathJax.

For those who may be interested, I will attach a PDF file of the same web site.

#### Attachments

• space-station-catch.pdf
577.9 KB · Views: 187
LAP3141 said:
OK. The velocity components are not the same in both frames.

After finding some more advanced descriptions on the web I am beginning to see the reason more clearly.

But the method I described is also NOT correct because it assumes that the velocity components are the same.

I started all this after discovering these related web pages:

http://www.phermi.com/space-station-catch
http://www.phermi.com/space-station-catch-solution

In the solution given, the velocity components in the rotating frame are transformed into the stationary frame and these transformed components are used to determine the trajectory in the stationary frame. Then the stationary frame trajectory is transformed back into the rotating frame.

This procedure is a bit awkward and I am trying to learn a more systematic approach.

But the solution which I described does NOT produce the curves shown in the solution web page. This is due to the assumption that the velocity components are equal in both frames.

I will study some more advanced web pages until I get the correct mathematics.

Velocity components are the same only if the frames are at rest with respect to each other.

PeroK said:
Velocity components are the same only if the frames are at rest with respect to each other.

I think the problem I was having is that intuitively it seemed to me that the INITIAL velocity components, at the t=0 instant, should be same in both frames.

In this case, the particle is shot from the rotating disc. But the same situation would occur if the particle were shot from the stationary frame instead.

LAP3141 said:
But the method I described is also NOT correct because it assumes that the velocity components are the same.
I thought that you were transforming the positions (trajectory). The method you describe is correct for that, not for transforming velocities.

Dale said:
so either you made an algebra mistake or your expectation of what it should be is wrong.

This response is just an aside, but I am using wxmaxima to do the calculations so there should be no algebra errors:

https://wxmaxima-developers.github.io/wxmaxima
Computer algebra systems eliminate those all too common pencil-pushing errors and the blame can then only be placed on the conceptual understanding of the person doing the calculations.

Dale
LAP3141 said:
Computer algebra systems eliminate those all too common pencil-pushing errors
I agree. This is a good approach, although you can still make button-pushing errors, but usually those are easier to catch.

LAP3141 said:
I think the problem I was having is that intuitively it seemed to me that the INITIAL velocity components, at the t=0 instant, should be same in both frames.

In this case, the particle is shot from the rotating disc. But the same situation would occur if the particle were shot from the stationary frame instead.

Relative motion is relative motion. ##t= 0## is just an arbitrary label. If the relative velocity at ##t =0## were zero, then that would be different.

In general ##u'= u -v##, where ##u'## is the velocity in a frame moving at velocity ##v##.

PeroK said:
In general ##u'= u -v##, where ##u'## is the velocity in a frame moving at velocity ##v##.
In this case the velocity of the moving frame, v, would be the instantaneous tangential velocity, w*R, of the rotating frame at the initial coordinates at t=0.

Is that correct?

LAP3141 said:
In this case the velocity of the moving frame, v, would be the instantaneous tangential velocity, w*R, of the rotating frame at the initial coordinates at t=0.

Is that correct?

Actually, with a rotation it's more complicated than that. The Cartesian unit vectors are changing between frames over time.

PeroK said:
Actually, with a rotation it's more complicated than that. The Cartesian unit vectors are changing between frames over time.

For a constant velocity the transformation seems quite simple.

Mathematically, the relation between the velocity in the two frames is:

v_inertial = v_rot + w x r

where "x" is the cross product operator and w the angular momentum vector.

Basically, v_rot is determined and then transformed into the inertial frame where it is used to calculate the inertial frame trajectory. Then the inertial frame trajectory, a straight line, is transformed back into the rotating frame.

It works. I have attached a wxmaxima file, exported as PDF, that shows this method as well as a graph of the rotating frame trajectory.

But beware. The wxmaxima is still incomplete because it was written when I did not fully understand what was happening. I tried to generalize things to accommodate any initial position and velocity.
Input cell #5 represents the transformation of velocity from rot to inert. I should have added the complete derivation. If anyone is interested I can provide a more complete wxmaxima file with comments later.

#### Attachments

• space-station-catch-wxmaxima.pdf
602.1 KB · Views: 192
Dale

## 1. What are rotational frame transformations?

Rotational frame transformations are mathematical equations used to convert coordinates and vectors from one reference frame to another that is rotating at a constant angular velocity.

## 2. Why are rotational frame transformations important?

Rotational frame transformations are important because they allow us to analyze and understand the motion of objects in different reference frames, which is crucial in fields such as physics, engineering, and astronomy.

## 3. How do rotational frame transformations work?

Rotational frame transformations involve using rotation matrices and angular velocity vectors to transform coordinates and vectors from one frame to another. The equations take into account the angle and direction of rotation, as well as the time elapsed.

## 4. What are some applications of rotational frame transformations?

Rotational frame transformations have many applications, including analyzing the motion of objects in rotating systems, predicting the behavior of celestial bodies, and designing rotating machinery.

## 5. Are there any limitations to rotational frame transformations?

While rotational frame transformations are a powerful tool, they do have limitations. They assume that the rotation is constant and that there are no external forces acting on the system. Additionally, they may not accurately describe the behavior of objects in highly non-inertial frames.

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