Find Angle of Departure for Ball Pushed Off Semi-Dome

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The discussion centers on determining the angle of departure for a ball pushed off a frictionless semi-dome. It emphasizes that at the angle of departure, the normal force must equal the centripetal force, expressed mathematically as mg cos a = mv^2/R. This equality occurs because, at that moment, there is no net force acting on the ball in the direction of the dome, indicating it is on the verge of losing contact. The normal force acts perpendicular to the surface, while the centripetal force directs toward the dome's center. Understanding this balance of forces is crucial for solving the problem accurately.
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A ball of mass m is being pushed off the top of a frionctless semi dome by a small negligible force. Find the angle where the ball left the surface.

I know how to solve this problem, but I don't understand the concept behind it.

Why does the normal force have to equal the centripeal force at the angle of departure a??

ie mg cos a = mv^2/R
 
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Hyperreality said:
Why does the normal force have to equal the centripeal force at the angle of departure a??

ie mg cos a = mv^2/R
the normal force is the force that is perpendicular to the surface, in this case the dome, it is sliding on, the normal force is pushing it way from the surface.
the centripetal for is the force that is force that is directed down toward the center of the dome it is sliding on/around, the centripetal force is holding it onto the dome.

so what those forces are equal at thay instant, there is no acceleration, or ner force toward or away from the dome, a moment before and it would still be pressing up against the dome, and a moment after and it will be accelerating away from the dome.
at the instant the forces are equal, the object looses contact and then starts to move away once the force pushing it away becomes greater.
 
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