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Yesterday I was looking at an old sloan video that describes motion in inertial and non-inertial frame. An experiment was actually like this. Two persons are sitting on the opposite side of a table fixed to a turning platform. The platform is rotating in uniform circular motion. Now Guy1 pushes a ball over the frictionless surface of the table in a straight line towards Guy2. The question what will be the motion of the ball from a viewer inside the rotating frame and to someone outside in fixed frame of reference. I got little confused as to how to conceive the fictitious force. What will the motion be? In general, how to derive equation of motion in non-inertial frame. Please also add some good references on intuitive understanding of these type of problems.

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  • $\begingroup$ The problem is that forces exist innately in inertial frames only. By writing out a vector of a force/moment couple on a moving frame it no longer represents a force in the sense we understand exactly. It is somewhat different. $\endgroup$ Dec 29 '15 at 16:05
  • $\begingroup$ But D'Alembert's principle actually tries to do that--isn't it? $\endgroup$ Dec 29 '15 at 16:06
  • $\begingroup$ D'Alembert can convert a dynamic problem into a static one by converting inertial forces into equal and opposite static forces ($F+(-m a)=0$) but I think this principle breaks down on non-inertial frames. $\endgroup$ Dec 29 '15 at 16:11
  • $\begingroup$ Sounds like this video: archive.org/details/frames_of_reference If not, viewing it may be instructive... $\endgroup$
    – DJohnM
    Dec 29 '15 at 18:06
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The ball will move in a straight line according to an observer outside. But the observer is rotating, he will see ball deflecting away. This is called $\textbf{coriolis effect}$. Watch this video to understand properly. coriolis effect

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The motion for viewer inside rotating frame will be a straight line while for inertial frame observer or outside person it will be circular as motion is relative to the frame.

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  • $\begingroup$ This answer is exactly the opposite of the true behaviour... $\endgroup$
    – DJohnM
    Dec 29 '15 at 18:07

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