my question is,if we very slightly put a ball on smooth(frictionless) rotating disc then what would happen to ball. It will fall or not. because with respect to rotating disc the ball will be acted upon by two forces centrifugal and coriolis force because of which we conclude that it will fall ,specifically by following a curved path. BUT MY BIG CONFUSION here is that,with respect to ground(or any inertial) frame ,there is no centrifugal(pseudo)force acting and also no Coriolis (pseudo) force.there is no friction also to provide required centripetal force,then why would the ball fall when net force is coming 0 with respect to ground.And if it fall then what force is it that accelerates it from 0 velocity, at the instant when we put it,to non zero value that makes it moving and eventually fall?


closed as unclear what you're asking by user259412, AccidentalFourierTransform, Jon Custer, ZeroTheHero, rob Mar 20 '17 at 20:12

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  • $\begingroup$ If you gently put a ball on the disk, which with respect to you is at rest, then it moves uniformly in the non inertial disk frame, but remains exactly at rest in your frame. $\endgroup$ – Aritro Pathak Mar 20 '17 at 13:20
  • $\begingroup$ No, it will not fall $\endgroup$ – user126422 Mar 20 '17 at 13:26

I'm not sure of the geometry of the problem, and what you mean by the ball 'falling'. If you assume a frictionless horizontal disk that's rotating 'anti-clockwise' at some constant angular velocity $\omega\hat{z}$, and you put a ball gently on the disk at some radial distance $r$ from the center of the disk, then if in your inertial frame the ball is at rest, then in the rotating frame, the ball has a initial velocity $\vec{v_{rot}}=-\omega r \hat{\theta}$. Thus the initial Coriolis force on the ball is $ -2m (\omega\hat{z} \times (-\omega r \hat{\theta}))=-2m\omega^{2}r \hat{r}$. The initial centrifugal force is $-m(\vec{\omega} \times (\vec{\omega}\times \vec{r}))= m\omega^{2}r \hat{r} $.

Thus in the non inertial rotating frame, the net initial force on the ball is $-2m\omega^{2}r \hat{r}+m\omega^{2}r \hat{r} = -m\omega^{2}r \hat{r} $. This is precisely the force that produces uniform circular motion, and thus the ball moves at the same constant tangential speed $\omega r$, and at all subsequent stages of motion, the force on the ball remains exactly the same.

This makes sense, because in your inertial frame, the ball was initially at rest, and in the absence of friction there is no horizontal force on it, and hence it remains at rest. With respect to the uniformly rotating disk, the ball keeps performing uniformly circular motion in the reverse direction, as we saw.


If the contact is indeed frictionless then the only force acting on the ball is the normal reaction which keeps it on the plane of the disk. The ball will not move radially or tangentially. Remember, Newton's laws only work with inertial reference frames.

Things get much more interesting if friction is involved. This is because initially the ball is going to slip and the resulting motion is far more complex than imagined. For certain combinations of disk spin and friction the ball is going to start moving inwards first before it transitions to pure rolling and eventually falling off.

I have done simulations of smooth balls on rotating disks and was very surprised by the results.

  • $\begingroup$ wow! may I ask how easy or difficult it is to perform those simulations? $\endgroup$ – Aritro Pathak Mar 20 '17 at 17:45
  • $\begingroup$ @AritroPathak Depends on the software you use for the simulations. $\endgroup$ – ja72 Mar 20 '17 at 17:48

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