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In general, I know that if you have a vector $\vec{F}$ in a rotating frame, and the frame has an angular velocity $\vec{\Omega}$ that the time derivative of $\vec{F}$ in a fixed frame would be $$\frac{d\vec{F}}{dt}=\left(\frac{d\vec{F}}{dt}\right)_r+\vec{\Omega}\times\vec{F}.$$

However, I'm confused how or if this would change if there are multiple angular velocities attached to a rotating axis. Let's say our rotating frame is as below. Angular Velocities

This angular velocity $\vec{\Omega_{z'}}$ has its own angular velocity $\vec{\Omega_y}$. My original thoughts are to simply combine the angular velocities into a single vector $\vec{\Omega_T}=\vec{\Omega_y}+\vec{\Omega_{z'}}$, but since the axis $z'$ is moving I'm not sure if it's that simple.

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  • $\begingroup$ There are not multiple angular velocity vectors here. There's only one, in this case represented by $\vec \Omega = \Omega_y \hat y + \Omega_x \hat x$. How one represents angular velocity (the map) and what angular velocity is (the territory) are two different things. Always beware of confusing the map for the territory. $\endgroup$ – David Hammen Jul 27 '18 at 19:21
  • $\begingroup$ Should I change the title of my question? $\endgroup$ – WnGatRC456 Jul 27 '18 at 19:47
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    $\begingroup$ Related : Velocity in a turning reference frame $\endgroup$ – Frobenius Jul 27 '18 at 20:13
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As was mentioned in the comments, there is only one angular velocity $\vec{\Omega}_T=\Omega_y\hat{y}+\Omega_{z'}\hat{z'}$. This is confirmed here from some MIT lecture notes. It seems my intuition was correct. EDIT: If you want to use this to find the velocity of a vector, you need to cast this into the global frame first.

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