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I was going through HC Verma's lecture on this and he mathematically derived the pseudo force in reference frames that are going under translational acceleration with respect to an inertial reference frame. But according to me you should be able to apply the same derivation for rotating reference frames, but apparently you cant, and I cant figure out why.

https://youtu.be/tIL7TuT--BY?list=PLWweJWdB_GuLYUIxyXQswOt6JEPhphrrR&t=653

Watch from 10:53 to 16:30 he derives it. shortly after he tells this cant be applied for rotational. I cant see where the mathematics breaks down. $\overrightarrow{r_{ps_0}}=\overrightarrow{r_{ss_0}}+\overrightarrow{r_{ps}}$. This is true in general regardless of the frame. now we can simply double differentiate wrt to time on both sides and obtain same equation he did with translational acceleration. but this is incorrect where am I going wrong?

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  • $\begingroup$ What exactly do you mean by can't be applied for rotation? in reference frames with non zero torque, we can apply the concept of pseudo torque. $\endgroup$
    – Dev Not Taken
    Commented Feb 4 at 14:28
  • $\begingroup$ I mean the equation he derived in the end can't be applied when the reference frame is rotating. i.e the psedo force does not equal ma. where a is the acceleration of the frame $\endgroup$
    – Hammock
    Commented Feb 5 at 0:58
  • $\begingroup$ If you're talking about a rotating frame, then the pseudo force is called Coriolis force. $\endgroup$
    – Darth Vader
    Commented Feb 5 at 5:14
  • $\begingroup$ ik but what I am saying is why doesn't this derivation apply there $\endgroup$
    – Hammock
    Commented Feb 5 at 7:42

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enter image description hereIt's because in case of rotational frames of referance at each position the translational acceleration(a) is different and also there will be a psudo force the centrifugal force . This centrifugal force acts in oppositie direction of radius, magnitude equal to centripetal force at the point on which object is. As a result what you have is a continuosly changing acceleration which is why you can't generalise the psuedo force like we did in case of translational

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
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    Commented Feb 6 at 12:28

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