In the following question:
A puck is placed on a friction less disk that is rotating with angular velocity $\vec \omega$. What is the equation of motion with respect to the rotating frame of the puck?
The contradiction that I can not resolve is that it seems really obvious that the puck will rotate with angular velocity $ - \vec \omega $ in the frame of the disk. But if look at the puck from the frame of reference of the disk, we can see that there is a pseudo centrifugal force acting on the puck. This is the only force that acts on the puck in this frame.
Hence, the puck should accelerate radially outward in this frame...but this is an exact opposite to the expected answer.
Answer
The centrifugal force is not the only one which acts on the puck in the rotating frame, there is also the Coriolis force. The sum of the two is given by $$ \vec{F} = -m \vec{\omega} \times (\vec{\omega}\times\vec{r})-2m\vec{\omega}\times\vec{v} $$ but in the rotating frame the velocity of the puck is $$ \vec{v}=-\vec{\omega}\times\vec{r} $$ and substituting we find $$ \vec{F} = m \vec{\omega} \times (\vec{\omega}\times\vec{r}) $$ which is equal and opposite to the centrifugal force, and is exactly the force needed for the observed circular motion of constant speed.
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