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If we have a ball in a frictionless tube, and we begin to spin the tube about one end, the ball will exit the tube, correct? Indeed, there must be some centripetal acceleration, but then what force is causing the centripetal acceleration? The tube's normal force only acts in a perpendicular direction, so what force could possibly be causing the ball to rotate?

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  • $\begingroup$ The ball will not rotate - it will slide out of the tube. $\endgroup$ – user27118 May 24 '15 at 3:08
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The ball will fall out of the tube, but I think this is precisely because there is no friction to supply the centripetal acceleration. There is still a normal force however this is not towards the point of rotation. Friction would supply a force along the direction of the tube. Since you cannot supply an adequate centripetal acceleration, the ball should fall out.

Remember, you must provide a centripetal acceleration to keep it in circular motion. If you don't have that, then no circular motion! This is simply Newton's first law: the ball would "prefer" to move in a non-circular path and must be coerced not to.

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  • $\begingroup$ I totally agree with the fact that the ball mustn't rotate.But see a case where you would have an infinitely long tube. The force analysis does show the necessity of 2 forces, one being the Coriolis force, responsible for giving the necessary boost to the ball along the tangential direction of course, and the other being ,what looks like having the same magnitude and direction as a centripetal acceleration would have! Any thoughts on what might be providing the later, given that only direction in which the tube can interact with ball being the tangential direction?thanks. $\endgroup$ – Goutham Dec 16 '16 at 19:06

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