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Consider an inertial reference frame $I$, and a point object $K$ which is at rest with respect to $I$. $K$ is not at the origin of $I$.

Consider another reference frame $I’$, with the same origin as $I$, but rotating along a specified axis with angular velocity $\omega$ with respect to $I$.

From the reference frame $I’$, the object $K$ appears to be revolving around the origin of $I’$ with angular velocity $\omega$.

In the inertial frame $I$, no external force was needed on the body $K$ to describe its motion.

On the other hand, in the non-inertial frame $I’$, we need a pseudo-force to describe the body using Newtonian laws. Since in this frame, the body is revolving, so the required pseudo-force is centripetal.

However, in a uniformly rotating non-inertial frame, the effective pseudo-force should be centrifugal, instead of centripetal. So we have a contradiction.

I am sure I am missing something, but what?

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If $K$ would be at rest in $I'$ then one would indeed observe a centrifugal force from $I'$. From $I$ one would see a real force that makes $K$ co-rotate (such as friction).

If $K$ is now at rest in $I$, then from $I'$ you would see that $K$ has a velocity $\mathbf{v}$ in the azimuthal direction. The fictitious force acting on $K$ is then the Coriolis force $\mathbf{F}_{\text{Coriolis}} = -2m\boldsymbol{\omega}\times\mathbf{v}$. You will find that it points in the negative radial direction.

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  • $\begingroup$ So the Coriolis force provides $2m\omega^2r$ radially inwards and the centrifugal force provides $m\omega^2r$ radially outwards, right? So, effectively we get a centripetal supply of $m\omega^2r$? $\endgroup$
    – Soham Saha
    Commented Jun 11 at 16:21
  • $\begingroup$ Actually, I had not learnt about the Coriolis force before. Thanks! $\endgroup$
    – Soham Saha
    Commented Jun 11 at 16:24
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    $\begingroup$ Yes in this case, if we use cylindrical coordinates, $\boldsymbol{\omega} = \omega\hat{\mathbf{z}}$ for example and $\mathbf{v}$ is purely azimuthal (in the negative $\phi$-direction). So with the RHR you would find that $\mathbf{F}_{\text{Coriolis}}$ is purely in the negative $r$-direction with a magnitude $2m\omega v$. $\endgroup$
    – Jesse
    Commented Jun 11 at 17:14

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