# Time derivative of vector in rotating frame with angular velocity about a rotating axis

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. 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.

• 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. Jul 27, 2018 at 19:21
• Should I change the title of my question? Jul 27, 2018 at 19:47
• Jul 27, 2018 at 20:13

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.