Rolling ball in a enclosed cone problem

[MarkoM27]

Hello , I am new here , and at start id like to say that i`m lousy at formulas and math :) . I've been searching and googled my problem and i couldn't find any solution to it . So here it is. Ball and a cone are rubber coated .Ball is rolling inside enclosed cone, when ball reaches speed it will climb up to cone base, and continue to roll if speed is maintained. It will not fall down and roll on 2 friction points. Cone is grounded and stationary. Ball will continue to roll inside a cone but it will not be able to fall outside it because its closed at base . Id like to know mathematics and physics behind this motion. What effects has gravity on it and what forces are acting in this motion. I have one lousy picture i hope someone will post a solution to this problem . Cheers and question everything :) http://postimg.org/image/62o1tlrzl/

 

[A.T.]

Ball is rolling inside enclosed cone

A related problem, that might be modified to a cone, is the cylinder:



http://demonstrations.wolfram.com/RollingBallInsideACylinder/

http://www2.eng.cam.ac.uk/~hemh/ball_in_vertical_cylinder.htm

 

[gsal]

You need to visualize two points of interest, the point of contact between the ball and the cone and the center of mass of the ball.

You also need to visualize, say, x and y-axis like the Cartesian one, but place the x-axis parallel to the surface of the cone and let the origin be also at the center of the ball.

First, we have gravity acting on the center of mass of the ball...this force is pointing straight down but it can be represented by its two orthogonal components along the x-y axis. The x component (Gx) wants to roll the ball down along the inside wall of the cone. The y component (Gy) pushes it against the cone inner surface.

The there is centrifugal force acting on the center of mass of the ball...this force is horizontal and pulling radially out and can be represented by its two orthogonal components along the x-y axis. The x component (Cx) wants to roll the ball up along the inside wall of the cone; under the conditions described, this force equals Gx and, thus, the ball does not roll up or down. The y component (Cy) pushes the ball against the cone inner surface.

The cone needs to be strong enough to counter act Gy+Cy.

Hope this helps