Find Angle of Departure for Ball Pushed Off Semi-Dome

[Hyperreality]

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

 

[mrjeffy321]

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.

 

[GSXR750]

The normal force and the centripetal force are related in this scenario because they are both acting in the same direction, perpendicular to the surface of the semi-dome. In order for the ball to maintain a circular path as it rolls down the dome, the normal force must provide enough support to counteract the centripetal force.

At the angle of departure, the ball is just about to leave the surface of the dome and begin its downward motion. At this point, the normal force is equal to the weight of the ball (mg) multiplied by the cosine of the angle (cos a), which represents the component of the weight acting perpendicular to the surface. This normal force must also be equal to the centripetal force, which is given by mv^2/R, where v is the velocity of the ball and R is the radius of the circular path it is following.

Therefore, by setting these two forces equal to each other, we can solve for the angle of departure a. This tells us at what angle the ball will leave the surface of the dome and start its downward motion. It is an important concept to understand in order to accurately predict the trajectory of the ball and ensure that it follows a circular path as it rolls down the dome.