I have been studying fictitious forces, such as the centrifugal force and Coriolis force. The equation for the centrifugal force is given by: $$F_{centrifugal}=-m\omega\times(\omega\times r)$$ My question is this, what does $\omega$ represent? I can see three possible options for a situation where the origins of the inertial frame and non-inertial frame do not coincide?

  1. The angular velocity of the origin of the non-inertial frame about the axis of the inertial frame.

  2. The angular velocity of the axis of the non-inertial frame about its own origin.

  3. A combination of the above.

If it is 3, please can you explain how we combine them.

  • $\begingroup$ Hi, I wrote a quick explanation of how to link $\omega$ to the rotation of the "3 axis of the non inertial frame" wrt the "3 axis of the inertial frame" in this question (call it "option number 4", as in the answer of Emilio below): physics.stackexchange.com/q/630781/226902 $\endgroup$
    – Quillo
    Apr 20, 2021 at 14:29

1 Answer 1


Option 4, none of the above.

Your option 1 is wrong because points don't rotate. Your option 2 is closer to correct, but ultimately still wrong. You're overly hung up on points (the origin).

It might help to get a handle on what "rotation" is. Points don't rotate. Better said, a rotated point is indistinguishable from the original. What about one dimensional space? Rotating a line about itself doesn't change the coordinate of a point on that line one iota. Once again, rotation doesn't make quite sense here.

Two dimensional space is where the concept of rotation begins, and indeed, it helps to look at rotations in higher dimensional spaces as a composite of two dimensional rotations. Only one parameter ("angle") is needed to describe a rotation in two dimensional space. Rotation is not a two-vector in two dimensional space. Rotation is not a four-vector in four dimensional space. Six parameters are needed to describe rotations in four dimensional space, ten in five dimensional space.

Our three dimensional space is the only space where the number of parameters needed to describe a rotation is equal to the dimensionality of the space. This is one of the reasons why we can treat angular velocity in three dimensional space as if it was a vector. Another key reason is the concept of an axis of rotation. That this axis must exist is the key point of Euler's rotation theorem. Any sequence of rotations in three dimensional space can be described in terms of a single rotation about some axis by some angle. That axis specifies a direction and the rotation angle specifies a magnitude. Direction and magnitude: That's a vector!

One reason I said "option 4, none of the above" is that you appeared to be a bit hung up on the origin. The origin doesn't really matter. It might help if you visually making some coordinate system markers. It's easy. Something like these:

Imagine making a bunch of them. Next, go find a playground with a kid's roundabout:

Spread your markers about the roundabout. Put one dead center, put others elsewhere. Now give it a spin. Think of each of those markers as representing a coordinate system. The coordinate system at the center of the roundabout is undergoing pure rotation. All of the others are undergoing a combination of rotation and acceleration. The origin does matter when it comes to velocity and acceleration of the origin of the frame, but it doesn't matter when it comes to angular velocity. All of those reference frames share the same angular velocity vector.

  • $\begingroup$ Ok so I get it now in the situation you have described, but what if one of these coordinate systems (not at the centre) was also rotating around one of its own axes? What $\omega$ do we use then? $\endgroup$
    – user43487
    Oct 1, 2014 at 12:09
  • $\begingroup$ @Joseph -- Here's my take on your last question (tell me if I interpreted it incorrectly). Imagine gluing one of those markers to a hamster wheel. Put the hamster wheel on the roundabout, and then give both the hamster wheel and the roundabout a spin. What's the angular velocity of the hamster wheel centered frame with respect to the ground? Simple: Add the angular velocities, vectorially. $\endgroup$ Oct 1, 2014 at 12:27

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