Talk:Rotating reference frame

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Pick some bones out of this:

Consider two frames of reference, one rotating and the other not; the co-ordinates at which an event occurs as described by one are obtained by applying a rotation to the co-ordinates of that event as described by the other; the required angle of rotation varies linearly with time.

Chose a point on the axis of rotation as origin and use cylindrical polar co-ordinates; the two systems will then share an axial co-ordinate z and a radial co-ordinate r, each of which is a length. Ignore relativistic effects, so the two also share a time co-ordinate, t. Complete the systems of co-ordinates with an angle p in the non-rotating one, P in the rotating one; have P and p coincide at t=0, so that P = p -w.t (modulo whole turns) for some angular velocity w.

Measure angles, p or P, positive if they are in the same direction as the rotation, thereby making w positive. Looking along the z axis in one direction will see such angles as clockwise; looking along it the other way, as anti-clockwise. Let z increase as one moves along the axis in the former direction, decreasing in the latter.

To obtain cartesian co-ordinates, replace

   * r and p with: x = r.Cos(p), y = r.Sin(p)
   * r and P with: X = r.Cos(P), Y = r.Sin(P) 

(Aside: I use Cos and Sin as the names of functions ({scalars}: |{angles}), whose inputs have units of angle; in contrast, cos and sin are ({scalars}: |{scalars}), equal to the composites of Cos and Sin after ({angles}: radian.i ←i :{scalars}); i.e. cos and sin measure angles in radians.) Now,

   * Cos(P) = Cos(p-w.t) = Cos(p).Cos(w.t) +Sin(p).Sin(w.t)
   * Sin(P) = Sin(p-w.t) = Sin(p).Cos(w.t) -Cos(p).Sin(w.t) 

whence

   * X = x.Cos(w.t) +y.Sin(w.t)
   * Y = y.Cos(w.t) -x.Sin(w.t) 

This tells us how positions are related to one another in the two frames: we shall now look into how the velocity and acceleration of a body, as seen by the two frames, are related. To do this, take x, y and z as the co-ordinates of the body, hence functions of time, and see how the time-derivatives of X and Y depend on those of x and y [that of z will pass seemlessly between the frames and have no impact on the others, so I ignore it]. We obtain

   *

     dX/dt
         = (dx/dt).Cos(w.t) +(dy/dt).Sin(w.t) -x.Sin(w.t).w/radian +y.Cos(w.t).w/radian
         = (dx/dt).Cos(w.t) +(dy/dt).Sin(w.t) +Y.w/radian

   *

     dY/dt
         = (dy/dt).Cos(w.t) -(dx/dt).Sin(w.t) -(y.Sin(w.t) +x.Cos(w.t)).w/radian
         = (dy/dt).Cos(w.t) -(dx/dt).Sin(w.t) -X.w/radian

Applying differentiation once more,

   *

     ddX/dt/dt
         = (ddx/dt/dt).Cos(w.t) +(ddy/dt/dt).Sin(w.t) -(dx/dt).Sin(w.t).w/radian +(dy/dt).Cos(w.t).w/radian +(dY/dt).w/radian
         = (ddx/dt/dt).Cos(w.t) +(ddy/dt/dt).Sin(w.t) +2.(dY/dt).w/radian +X.w.w/radian/radian

   *

     ddY/dt/dt
         = (ddy/dt/dt).Cos(w.t) -(ddx/dt/dt).Sin(w.t) -((dy/dt).Sin(w.t) +(dx/dt).Cos(w.t)).w/radian -(dX/dt).w/radian
         = (ddy/dt/dt).Cos(w.t) -(ddx/dt/dt).Sin(w.t) -2.(dX/dt).w/radian +Y.w.w/radian/radian

This enables us to describe the acceleration of the body, as seen by the rotating frame, as a sum of three terms:

rotated actual acceleration obtained, from the components of the acceleration seen by the non-rotating frame, by the same transformation as we had to apply to the components of position; this will exactly equal the force acting on our body, as described by the rotating frame, divided by the body's mass. We can include the z-component of actual accelleration as a natural component of this term; it is unaffected by the rotation.

the Coriolis term which is 2.w/radian times the result of rotating the velocity of the body (as seen by the rotating frame) through a right angle in the sense which maps a purely Y-wards vector into a purely X-wards one. (The name comes from someone who famously documented this effect.) The axial component of the velocity is ignored in this term.

the centrifugal term: which is the body's position, ignoring axial component, scaled by the square of w/radian. (The name means flying away from the centre.)

If someone in the rotating frame attempts to describe the physics they see, they thus see force = mass.acceleration modified by, in effect, the addition of two forces which arise as artefacts of their choice of frame of reference. [Interestingly, modern physics is much more at ease with this than its pre-Einstein antecedents; general relativity is quite happy to let us use the rotating frame, and effectively regards the two artificial forces as part of gravitation - but I digress.] � Exploiting the vector outer product

In three dimensions (and, let me stress, specifically in three dimensions) there is a product operator (which actually depends on your metric) which combines two 3-vectors, antisymmetrically, to produce a third. Specifically, if the [x,y,z] components of vectors u and v are [a,b,c] and [e,f,g], respectively, then those of their outer product, u^v, are [b.g-c.f, c.e-a.g, a.f-b.e]. [This relies on x, y, z being orthonormal co-ordinates for our metric; along with a presumption that the rotation which takes the y axis to the x axis appears clockwise when viewed from a position with positive z co-ordinate.] Crucially, u^v is perpendicular to both u and v, applying any scaling to either u or v scales u^v to the same degree and v^u = -u^v.

Now, both the Coriolis and centrifugal terms in our transformed acceleration have zero component in the z-direction, i.e. parallel to the axis; furthermore, the Coriolis term is perpendicular to the transformed velocity's projection into the plane perpendicular to the axis - it is, consequently, perpendicular to the transformed velocity itself (which is the sum of its given projection and its component parallel to the axis; adding two vectors perpendicular to the Coriolis term yields another). So we should compare the Coriolis term with the outer product of the transformed velocity and some vector in the direction of the axis. Since the Coriolis term is also proportional to w/radian, we may as well use this as the scale of the axial vector.

So let W be a vector parallel to our axis, of magnitude w/radian (which has units 1/time), pointing in the direction of increasing z. In the non-rotating frame, let s be our body's position, v = ds/dt its velocity and a = dv/dt its acceleration, with components

   * [x, y, z]
   * [dx/dt, dy/dt, dz/dt]
   * [ddx/dt/dt, ddy/dt/dt, ddz/dt/dt] 

respectively. Let S, V and A, likewise, be the position velocity and acceleration seen by the rotating frame, with components

   * [X, Y, z]
   * [dX/dt, dY/dt, dz/dt]
   * [ddX/dt/dt, ddY/dt/dt, ddz/dt/dt] 

Then S^W has co-ordinates [Y,-X,0].w/radian, giving us V = v +S^W. Likewise, the Coriolis term in A is simply 2.V^W. Furthermore, (S^W)^W has co-ordinates [-X, -Y, 0].w.w/radian/radian; so we are able to write our transformation, above, simply as

   * v = V -S^W
   * a = A -2.V^W +(S^W)^W 

and the force that our rotating frame sees act on a body with velocity V at position S is then just that seen by the non-rotating frame with an added term 2.V^W +W^(S^W) per unit mass. �

Im attempting to expand this article using some of the ideas on this talk page. However, this stuff will need rewriting as it has been copied from a web page. Any help in rewriting, paraphrasing etc, is welcome.--Light current 16:16, 28 October 2005 (UTC)[reply]

Im finding this difficult to reword as its written so well already!--Light current 22:15, 31 October 2005 (UTC)[reply]

Equations need 'mathifying' Can anyone with experince do this please?--Light current 22:18, 31 October 2005 (UTC)[reply]

Had a go at replacing the first para with something a bit longer. Feel free to improve/revert/whatever. ErkDemon 05:13, 31 October 2005 (UTC)[reply]

Im not sure how this fits into this page:

Examples of rotating objects Some

I am not exactly sure on what has to go there atm, but products into nothing, or the bracets in the first formula in the acceleration part are just messed up. Really gotta be fixed…--Seberg 13:40, 18 December 2006 (UTC)[reply]

To be brutal, this page is a disaster. I would propose a total rewrite form the ground up by an expert. ObsessiveMathsFreak 22:58, 29 December 2006 (UTC)[reply]

I would suggest deletion of the page and redirect to Fictitious force. Brews ohare (talk) 18:36, 10 July 2008 (UTC)[reply]

A proposal has been made to merge and/or redirect this page to Fictitious force. Please join the conversation there. Rossami (talk) 21:20, 10 July 2008 (UTC)[reply]

I'm here from the featured article,

The sentrifugal force, for example, depends on both v_r and r_r, where r_r is taken to be the position in the rotating reference frame. Would it not be clearer to write these forces depending on either only stationary reference frame quantities and/or only rotating reference frame quantities? — Preceding unsigned comment added by Serpej (talk • contribs) 12:05, 8 October 2014 (UTC)[reply]

The above statement is not very clear, but keep in mind that the centrifugal and euler forces do not depend on the velocity in the rotating reference system. Apart from the rotation vector, they are fully determined by the location only. The Coriolis force does depend on the velocity in the rotational frame. −Woodstone (talk) 15:51, 8 October 2014 (UTC)[reply]

Where I went to school, we always used a lower case omega for angular velocity. The linked wikipedia page also says that the proper symbol is the lower case letter omega ω, or rarely an upper case omega Ω.

I believe this page would be a lot easier to read if we used the conventional symbol instead of the rarely used one. — Preceding unsigned comment added by 89.200.192.1 (talk) 07:03, 20 July 2015 (UTC)[reply]

An editor has identified a potential problem with the redirect Transport theorem and has thus listed it for discussion. This discussion will occur at Wikipedia:Redirects for discussion/Log/2022 January 27#Transport theorem until a consensus is reached, and readers of this page are welcome to contribute to the discussion. DVdm (talk) 22:37, 27 January 2022 (UTC)[reply]

I have a doubt, suppose there is a turntable rotating with constant angular velocity w… the table is frictionless and a particle is kept on the table . So now from the knowledge of rotating frame, the acceleration as seen from the rotating frame is always radially directed d2r/dt2 (in r^ direction). But now in this case there is no friction, how will the tangential coriolis force be nullified to yield LHS=RHS in rotating frames. When there is no friction. We are getting an acceleration in tangential direction ; everything I wrote is with respect to rotating frame — Preceding unsigned comment added by 49.205.135.234 (talk) 15:41, 25 April 2024 (UTC)[reply]

New threads belong at the bottom of the page. I have moved your post from the top, down to the bottom.

Your enquiry is not a particularly good one. If there is no friction between the rotating turntable and the particle, the particle will never move in response to the movement of the turntable - it will retain its original motion until it departs from the edge of the turntable. If it is placed on the turntable with zero velocity relative to the ground, it will remain motionless relative to the ground, and so will never depart from the edge of the turntable. Dolphin (t) 11:54, 27 April 2024 (UTC)[reply]