Answers to a Roulette problem

  • Thread starter AdultJason
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Hello,

Can you help by providing some answers to a Roulette problem I have.

The problem is that a roulette ball spins in the grove of a cone until the power pulling it down exceeds the force created by the push of the spin of the ball. Also, the diameter of the ball exceeds the thickness of the grove.

If the above is true, then does a relationship exist between where the ball leaves the grove, which is dependent on the force put into the spin as well as the starting point of the spin and a certain co-ordinate on the cone that the ball will cross as it rotates downwards?

Any help will be much appreciated.
 

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  • #2
Astronuc
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The time when the ball leaves the groove would depend in its original angular velocity. To fall out of the groove, the gravitational force pulling the ball down or out of the groove, must exceed the centripetal force keeping the ball in place.

The ball travels in opposition to the spin of the cone, so the ball must be decelerating at some rate.


So, [itex]\omega (t) = \omega_o + \alpha * t[/itex], where

[itex]\omega[/itex] = angular velocity of ball
[itex]\omega_o[/itex] = initial angular velocity
[itex]\alpha[/itex] = angular acceleration, which is negative (-) for deceleration

This equation implies that angular acceleration is constant.

If not then the deceleration becomes

[itex] \int^t_{t_o}{\alpha(t)} dt[/itex], and to may = 0.

The deceleration would depend on the resistance between the ball and the groove, and the relative speed between the ball and the groove. In rouloutte the wheel rotates in the opposite direction to the revolution of the ball.
 
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