Rotating reference frame: Difference between revisions

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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(so <math>z' = z</math> and <math>\frac{\operatorname{d} \theta}{\operatorname{d} t} =\equiv \Omega,</math> (sowhich implies <math>\theta(t) = \Omega t + \theta_0</math> for some constant <math>\theta_0,</math> where <math>\theta(t)</math> denotes the usual angle in the <math>x-y</math>-plane formed at time <math>t</math> by <math>\left(x', y'\right)</math> and the <math>x</math>-axis),
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> so take <math>\theta_0 = 0</math> or some other integer multiple of <math>2\pi</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>