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I read in HC Verma that if we are observing the motion of a body from a rotating frame and the body under the observation is NOT in motion with respect to our frame then the centrifugal force is the sufficient pseudo force for the analysis of the motion. But if the body under observation is in motion with respect to our frame than some extra pseudo forces other than centrifugal forces are required for the analysis of the motion of the object in our frame. Please explain this additional pseudo force's reason and details.

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    $\begingroup$ If you apply the transformation in full generality you get the centrifugal force and the coriolis force, as derived at: physics.stackexchange.com/q/383 $\endgroup$ – user12029 Jan 30 '14 at 20:23
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    $\begingroup$ Rotating frames also experience the Coriolis force and (if there is an angular acceleration) the Euler force. This is explained in nauseating detail on the wikipedia. $\endgroup$ – dmckee Jan 30 '14 at 20:24
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    $\begingroup$ Related: physics.stackexchange.com/q/68002 $\endgroup$ – joshphysics Jan 30 '14 at 21:04
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I show you a typical situation where the centrifugal force is not enough to explain the dynamics.

Suppose the reference frame $K$ is rotating around the $z$ axis of an inertial reference frame $K_0$ with $\omega = \Omega {\bf e}_z$ and $\Omega>0$ constant.

Consider a point of matter $P$ at rest in $K_0$ far from the origin. In $K$, that point is seen rotating with $\omega'= -\Omega {\bf e}_z$ (as the rotation is around that fixed axis we can identify the axes ${\bf e}_z$ and the corresponding ${\bf e}_z'$ at rest in $K$).

If $m$ is the mass of $p$, the motion in $K$ needs a centripetal force: $${\bf F}= -m \Omega^2 \vec{OP}\:.\quad(1)$$ This force can only be due to pseudoforces, as no real force acts on $P$ (or, equivalently, the sum of real forces acting on $P$ vanishes), since $P$ stays at rest in the inertial reference frame $K_0$. However, if only the centrifugal pseudoforce were present it would not be enough as it is: $${\bf f}_{centrifugal} = + m \Omega^2 \vec{OP}\:.$$ To fulfil (1) another force with inverse direction and double magnitude is necessary.

It is the Coriolis' (pseudo)force: $${\bf f}_{Coriolis} = -2 m \omega \times {\bf v}_P\:,$$ where $\omega$ is the angular velocity of $K$ with respect to $K_0$ and ${\bf v}_P$ is the velocity of $P$ in $K$. One has: $${\bf v}_P = -\omega \times \vec{OP} = - \Omega {\bf e}_z \times \vec{OP}\:.$$ Therefore: $${\bf f}_{Coriolis} = 2 m \Omega^2 {\bf e_z} \times ({\bf e}_z \times \vec{OP})= - 2m \Omega^2 \vec{OP}\:,$$ so that $${\bf f}_{Coriolis} + {\bf f}_{centrifugal} = -m \Omega^2 \vec{OP}$$ as requested by (1).

Dropping the requirement of uniform rotation, and assuming that $\omega$ of $K$ respect to $K_0$ is arbitrary and that $K$ may also translate with respect to $K_0$, the complete set of pseudoforces acting on a point $P$ in $K$ is given by the following added four terms (the third one is noting but the centrifugal force written into a more general form): $$- m {\bf a}_O - m\omega \times (\omega \times \vec{PO}) - 2m \omega \times {\bf v}_P - m \dot{\omega} \times \vec{OP}\:,$$ where ${\bf a}_O$ is the acceleration of the origin $O$ of $K_0$ computed in $K$, ${\bf v}_P$ is the velocity of $P$ in $K$, $\dot{\omega}$ is the time derivative of $\omega$, which is, as before, the angular velocity of $K$ in $K_0$.

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Another psuedo-force is the Azimuthal Force (Sometimes called Euler's force) it arises when there is angular accleration. It's vector equation is $$F_{Azimuthal}=\mathbf{r} \times \mathbf{\alpha}$$, where $\mathbf{\alpha}$ is the angular accleration vector.

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