Recently I'm learning rotating frame kinematics, but I have come up with a question that confuses me a lot.
Let's say a vector $P$ connects the origin and a fixed point in the rotating frame, so it is constant in the rotating frame (local frame, $LF$), so that $P_{x,LF}$, $P_{y,LF}$, and $P_{z,LF}$ never change. The coordinates of $P$ in the inertial frame ($IF$) can be expressed as:
$P_{IF} = R_{LF2IF}P_{LF} $
where $R_{LF2IF}$ denotes the rotation matrix from the local frame to the inertial frame.
Based on the above equation, the time derivative of $P_{LF}$ is:
$\dot{P_{IF}} = \dot{R_{LF2IF}}P_{LF} + R_{LF2IF}\dot{P_{LF}}$ ------------- (1)
According to the rotating frame kinematics Rotating Reference Frame, the time derivative of $P_{LF}$ is: $\frac{dP_{LF}}{dt} = (\frac{dP_{LF}}{dt})_{LF} + \omega\times P_{LF}$
Because $P_{LF}$ is constant in the local frame, therefore: $\frac{dP_{LF}}{dt} = 0 + \omega\times P_{LF} = \omega\times P_{LF}$
However I have seen in many places where people consider $\dot{P_{LF}}$ to be zero, which makes me rather confused. It seems that the following condition is true:
$\frac{dP_{LF}}{dt} = (\frac{dP_{LF}}{dt})_{LF}$
But, why is this, why the cross product is not calculated in this case? When should the cross product be used and when should not?
Many thanks to anyone who would kindly help me to clear my confusion.
The cross product only appears when you take the time derivative of a vector in a rotating frame. In your case, the vector $P$ is constant in the rotating frame, so its time derivative is zero.
In general, if you have a vector $A$ that is constant in the rotating frame, then its time derivative in the inertial frame is given by:
$\frac{dA_{IF}}{dt} = \dot{R}_{LF2IF}A_{LF} + R_{LF2IF}\frac{dA_{LF}}{dt}$
where $\dot{R}_{LF2IF}$ is the time derivative of the rotation matrix. If you plug in the expression for $\frac{dA_{LF}}{dt}$ from the rotating frame kinematics, you will get equation (1) above.