# Homework Help: Primed, Unprimed, and Rotating

1. Apr 16, 2014

1. The problem statement, all variables and given/known data

So let's say I have a particle moving at some given latitude λ on or near the earth's surface, northen hemisphere (see diagram attached). Look at the coordiate system I have x',y'and z' with origin on the surface...and the velocity I'm given has components along those primed axis. So I'm given three components,
$$\vec V' = \vec V_{x'} + \vec V_{y'}+ \vec V_{z'}$$

I want to find Coriolis Force

In my text there is discussion about "primed coordinate system rotating with respect to an unprimed, fixed, inertial one."

I can't tell which one I should regard as being "fixed." If I treat xyz in my diagram as fixed, then x'y'z' is a "translated" coordinate system that is orbiting the z axis of the fixed system. But this is not the same as "rotating," as I understand it...and I don't think that would make any sense, so I am wondering if I should treat the x'y'z' system, which has origin at the surface, as stationary "inertial" and the xyz as the rotation system the angular velocity vector, which is defined as the vector $$\vec \omega = \omega \hat n$$ with n in direction of axis of rotation would therefore be in the z direction...

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2. Apr 16, 2014

### electricspit

I think it's much easier if you treat $(x,y,z)$ as the inertial frame. The primed coordinates are usually used for the non-inertial frame. The non-inertial frame is the frame which is moving (or in this case stuck to the surface of the Earth as it rotates).

Since the Earth is rotating, the frame in which you are observing the Earth rotating from a distance is the inertial frame. Does that make sense? I had trouble with this last year when I was in my mechanics class as well.

3. Apr 24, 2014

Yes, I figured that out as well. Treating problems like this is actually very systematic if you follow the derivations transforming all the dynamic variables for translating and/or rotating systems and use them accordingly.

I didn't do too well on this content though on the exam, but at least I got the Lagrangian stuff.

4. Apr 25, 2014

### D H

Staff Emeritus
If that's a professionally-drawn diagram, or even a professorially-drawn diagram, it is not that good. It looks to me like x', y', and z' point in random directions. (If it's your own diagram, its probably because you're confused. No insult intended.)

I don't know your text, but here's a good local coordinate system: East-North-Up. Stand facing north. Point your thumb of your right hand to the east, your index finger to the north, and your middle finger up. A nice right handed coordinate system. (Physicists and others have beat senseless the silly geologists who used to insist upon using left handed North-East-Up coordinates. It's East-North-Up or North-East-Down.)

*You* are the origin of this coordinate system. You yourself are rotating about the Earth's axis, at a good distance from it. That means this is an accelerating coordinate system. If you think about it for a bit, you should see that it is also a rotating coordinate system. Where your local east, north, and up axes point in inertial space depends on your latitude, your longitude, and on time of day.

There's an easy way to fix the acceleration part. Consider a third frame of reference whose axes point in the same way your east, north, and up axes point but whose origin is at a fixed east-north-up offset of {0,0,-RE} relative to you. In other words, this third frame has its origin at the center of the Earth. This frame is not accelerating. It's just rotating. Since it's a fixed offset from your position (as far as you are concerned), the velocity vector {vx', vy', vz'} in your frame remains unchanged in this third frame.

See if you can take it from here.

5. Apr 30, 2014

I just drew that diagram in paint so it's just crude but I figured it'd be easy to discern what I meant, the axis are mutually perpendicular and the xy plane is tangent to the surface.

That's tricky, I don't see that right-away. I understand that "I" am an accelerating frame since I'm traversing a curved path...I was thinking of this as an orbit, but I guess it could be a rotation as well but I'm not sure I'm reasoning this out correctly:

....If I imagine a normal line drawn from the surface, northern hemisphere say...the line will form some angle with the earth's axis of rotation, and if you image the earth rotates 180 degrees there will be a component of the normal line pointing in the opposite direction as on the other side....so in this sense it has actually rotated 180 degrees.... so the axis of rotation for this system would just be the along the projection of the earth's rotation onto the normal line?

My text did in fact define a "rotation" vector taking the ω and throwing on a unit vector in directed along the axis of rotation.... so maybe I just figured out why.

6. May 1, 2014

### D H

Staff Emeritus
The east-north-up frame at some point on the surface of the Earth is neither moving nor rotating when viewed from the perspective of what astronomers call an "Earth-centered, Earth-fixed frame", a frame whose origin is at the center of the Earth and whose axes rotate with the rotating Earth. (Note well: This use of "fixed" is opposite the meaning used by physicists, where "fixed frame" is a synonym for an inertial frame.)

Since that east-north-up frame is stationary in a frame that rotates with the Earth, it too is a rotating (but also translating and accelerating) frame.