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=== Relation between positions in the two frames ===
To derive these fictitious forces, it's helpful to be able to convert between the coordinates <math>\left(x', y', z'\right)</math> of the rotating reference frame and the coordinates <math>(x, y, z)</math> of an [[inertial reference frame]] with the same origin.<ref group=note>So <math>x', y', z'</math> are functions of <math>x, y, z,</math> and time <math>t.</math> Similarly <math>x, y, z</math> are functions of <math>x', y', z',</math> and <math>t.</math> That these reference frames have the same origin means that for all <math>t,</math> <math>\left(x', y', z'\right) = (0, 0, 0)</math> if and only if <math>(x, y, z) = (0, 0, 0).</math></ref> If the rotation is about the <math>z</math> axis with a constant [[angular velocity]] <math>\Omega,</math> meaning <math>\frac{\operatorname{d} \theta}{\operatorname{d} t} = \Omega</math> (so <math>\theta(t) = \Omega t + \theta_0</math> for some constant <math>\theta_0,</math> where <math>\theta</math> denotes the usual angle in the <math>x-y</math>-plane) and if the two reference frames coincide at time <math>t = 0</math> (meaning <math>\left(x', y', z'\right) = (x, y, z)</math> when <math>t = 0</math>), the transformation from rotating coordinates to inertial coordinates can be written
<math display=block>x = x'\cos(\theta(t)) - y'\sin(\theta(t))</math>
<math display=block>y = x'\sin(\theta(t)) + y'\cos(\theta(t))</math>
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Introduce the unit vectors <math>\hat{\boldsymbol{\imath}},\ \hat{\boldsymbol{\jmath}},\ \hat{\boldsymbol{k}}</math> representing standard unit basis vectors in the rotating frame. As they rotate they will remain normalized. If we let them rotate at the speed of <math> \Omega(t) </math> about an axis <math>\boldsymbol {\Omega}(t)</math> then each unit vector <math>\hat{\boldsymbol{u}}</math> of the rotating coordinate system abides by the following equation:
:<math> \frac{\mathrm{d}}{\mathrm{d}t}\hat{\boldsymbol{u}}=\boldsymbol{\Omega}(t) \times \boldsymbol{\hat{u}} \ .</math>
Then if we have a vector function <math>\boldsymbol{f},</math>
:<math> \boldsymbol{f}(t)=f_x(t) \hat{\boldsymbol{\imath}}+f_y(t) \hat{\boldsymbol{\jmath}}+f_z(t) \hat{\boldsymbol{k}}\ , </math>
and we want to examine its first derivative we have (using the [[product rule]] of differentiation):<ref name=Lanczos>{{cite book |url=https://books.google.com/books?num=10&btnG=Google+Search |title=The Variational Principles of Mechanics |author=Cornelius Lanczos |date=1986 |isbn=0-486-65067-7 |publisher=[[Dover Publications]] |edition=Reprint of Fourth Edition of 1970 |no-pp=true |pages=Chapter 4, §5}}</ref><ref name=Taylor>{{cite book |title=Classical Mechanics |author=John R Taylor |page= 342 |publisher=University Science Books |isbn=1-891389-22-X |date=2005 |url=https://books.google.com/books?id=P1kCtNr-pJsC&pg=PP1}}</ref>
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:<math>\mathbf{v} \ \stackrel{\mathrm{def}}{=}\ \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}</math>
The time derivative of a position <math>\boldsymbol{r}(t)</math> in a rotating reference frame has two components, one from the explicit time dependence due to motion of the particle itself, and another from the frame's own rotation. Applying the result of the previous subsection to the displacement <math>\boldsymbol{r}(t),</math>
:<math>
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When the expression for acceleration is multiplied by the mass of the particle, the three extra terms on the right-hand side result in [[fictitious force]]s in the rotating reference frame, that is, apparent forces that result from being in a [[non-inertial reference frame]], rather than from any physical interaction between bodies.
Using [[Newton's laws of motion|Newton's second law of motion]] <math>\mathbf{F}=m\mathbf{a},</math>
* the [[Coriolis force]]
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==References==
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{{reflist|group=note}}
==External links==
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