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## Main Question or Discussion Point

Imagine two frames one inertial (x,y,z) and the other rotating (x',y',z'), their origins are always coincident. The rotating frame is rotating as seen from the inertial frame with a time-dependent angular velocity ##\boldsymbol{\Omega}(t)=(\Omega_x(t),\Omega_y(t),\Omega_z(t))##. In the rotating frame, a particle is moving in a circular fashion about the rotating frame's origin in the x-y plane at a radius ##r## in the form ##\boldsymbol{r}'(t)=r(\cos(\phi(t)),\sin(\phi(t)),0)##, where ##\phi(t)## is the angle between the position vector ##\boldsymbol{r}'(t)## in the rotating frame and the x' axis and ##r=|\boldsymbol{r}'|##.

I would like to know what the velocity of the particle is as seen from the inertial frame at time t, I have started this by considering the relation between velocities of particles in two frames, one inertial the other rotating

\begin{equation}

\frac{d\boldsymbol{r}}{dt}=\frac{d\boldsymbol{r}'}{dt}+\boldsymbol{\Omega}\times\boldsymbol{r}'

\end{equation}

Where the term on the R.H.S is the velocity of the particle as seen from the inertial frame, and the first term on the L.H.S is the velocity of the particle as seen from the rotating frame, the second term being the rotation of the rotating frame as seen from the inertial frame.

I have some doubts about using this though, Eq.(1) seems to be an instantaneous velocity, it doesn't describe the evolution of the velocity as seen from the inertial frame for all times t. For example, if I give the position vector ##\boldsymbol{r}'## some time dependence ##\boldsymbol{r}'(t)## substitute into Eq.(1), what I get is the instantaneous velocity of the particle as if I started at the position ##\boldsymbol{r}'(t)##? And similarly, giving a time dependence to ##\boldsymbol{\Omega}##.

Or have I missed understood this equation?

I would like to know what the velocity of the particle is as seen from the inertial frame at time t, I have started this by considering the relation between velocities of particles in two frames, one inertial the other rotating

\begin{equation}

\frac{d\boldsymbol{r}}{dt}=\frac{d\boldsymbol{r}'}{dt}+\boldsymbol{\Omega}\times\boldsymbol{r}'

\end{equation}

Where the term on the R.H.S is the velocity of the particle as seen from the inertial frame, and the first term on the L.H.S is the velocity of the particle as seen from the rotating frame, the second term being the rotation of the rotating frame as seen from the inertial frame.

I have some doubts about using this though, Eq.(1) seems to be an instantaneous velocity, it doesn't describe the evolution of the velocity as seen from the inertial frame for all times t. For example, if I give the position vector ##\boldsymbol{r}'## some time dependence ##\boldsymbol{r}'(t)## substitute into Eq.(1), what I get is the instantaneous velocity of the particle as if I started at the position ##\boldsymbol{r}'(t)##? And similarly, giving a time dependence to ##\boldsymbol{\Omega}##.

Or have I missed understood this equation?