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Suppose I have circular, frictionless disk, and in the center of the disk there is a small box. The disk is then rotated. I gently push the box to disorient it from the center. Will the box accelerate out of the frictionless disk due to pseudo(centrifugal) acceleration, or will it go out with the constant velocity with which I pushed it?

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Since after the push there are no forces acting on the box parallel to the surface of the disk, the box will just continue in a straight line at a constant velocity. We know this must be true from Newton's first law. Although, this is if you are working in an inertial frame of reference where centrifugal forces, or any other pseudo-forces, are not present.

If you are working in the frame rotating with the disk, you would find that the centrifugal and Coriolis forces would combine in such a way where you will see the box spiraling outwards with the radial coordinate increasing at a constant rate. But these pseudo-forces are present because of working in a non-inertial reference frame. It has nothing to do with the fact that the disk is spinning. The fact that the disk is spinning does nothing to influence the box.

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There minsconception you have is about when pseudo force of centrifugal or centripetal act, since you push the rotating block it will continue to move in a straight line, there is no force parallel to the direction in which block is moving.

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Will it accelerate out of the disk

It will not "accelerate" out of the disk, just slowly, constantly in its own pace with its constant speed move out away from the centre until it falls off of the disk edge.

It will not be stopped, since there is no friction. And it will not accelerate any further after your push is done, since there are no other unbalanced forces. Horizontal forces are necessary for horizontal acceleration due to Newton's 2nd law, and there are no such forces present.

And this is all that will happen. No need to think of nonexisting "centrifugal forces" or "centrifugal acceleration" (although I'm not sure what that is).

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