Questions about a particle shot off of a rotatiting disc.

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The discussion centers on the dynamics of a particle dropped onto a rotating disc with angular velocity \( w \). When the particle is released at a radius of \( 0.5r \), the centripetal acceleration is calculated as \( A_c(0.5r) = \frac{rw^2}{2} \). However, if the particle is dropped directly onto the disc, it does not exhibit any motion due to the assumption of negligible friction, resulting in no circular motion. The key conclusion is that without friction, the particle remains stationary upon contact with the disc.

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Say you have a disc that rotates with angular velocity w. Assume that you know the value of w and the radius of the disc. You drop a particle on the disc while it is rotating half way out from the center of the radius. So the the centripetal acceleration on the particle is

Ac(@.5r) = rw^2/2

How can you go about finding how long in terms of rotations will it take the particle to be shot off the disc ignoring friction. This is not a specific problem just somthing I have been thinking about.
 
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first of all, the accerlation the ball feels if it were to move in circular motion with radius (r/2) would be a = 2rw^2

second, if you drop ball directly onto the disc, ie. the trajectory of the fall is perpendicular to the plane of the disc, then the ball would not move, because you told us to ignore friction.

In other words, the ball wouldn't exhibit circular motion, let alone motion at all.
 

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