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-   -   Roulette F=mv^2/r (http://www.physicsforums.com/showthread.php?t=235816)

a.mlw.walker May18-08 09:49 AM

roulette F=mv^2/r
 
Hey Everyone,

I'm a first year mech eng student and have been studying circular movement.
Gonna state a few things, please pick me up where i'm incorrect:

A roulette wheel. The ball spins around, and is held against the side of the wheel due to F=mv^2/r

If m and r are constant, then whatever speed the ball starts spinning at, will always fall off the edge of the wheel at the same speed.

Its speed at a moment is 2PI/time taken for one revolution - angular velocity?

This is where i am maninly confused:

Its acceleration is change in two measured velocities/difference in time taken for those velocities? i.e change in speed over change in time? This deceleration is a constant until the ball falls from the rim.

The ball will only stay against the rim if v^2/r > g from:

f = mv^2/r = mg

when they are equal the ball is about to fall but not quite, and the m's cancel.

v^2/r is centrepetal acceleration. what affect does this have - how can i think about how this affects the ball?

if a ball has a velocity of 10m/s and a deceleration of 0.1m/s/s then the ball will take 100 seconds to stop?

How far off am i?


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