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Roulette F=mv^2/r

  1. May 18, 2008 #1
    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?
  2. jcsd
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