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Rotational Kinematics Question

  1. Dec 10, 2009 #1
    1. The problem statement, all variables and given/known data
    The wheel of fortune wheel, whose radius is 3.00m, has 20 different numbers, at equal intervals, for a contestant to land on. The second contestant can spin the wheel, and wishes to land on the $1.00. The wheel starts off 4 positions away from the $1.00. Assuming a deceleration of 0.400 rad/s2, what must be the minimum tangential velocity for the contestant to reach the $1.00, if they pull the wheel down?

    2. Relevant equations
    2pi rads is a full circle
    w^2 = wo^2 + 2(alpha)(theta)
    alpha= angular acceleration
    theta is distance in radians travelled
    w^2 is final angular velocity squared
    wo^2 is initial angular velocity squared
    Tangential Velocity = Radius x w

    3. The attempt at a solution
    2pi/20= 0.314 rads per interval
    the goal is 4 intervals away from the start therefore 4x0.314= 1.257 rads travelled.
    it wants to land 4 intervals away from the start so the final angular velocity = 0 rad/s
    therefore w^2 = wo^2 + 2 (alpha)(theta)
    0 = wo^2 + 2(-.4rad/s^2)(1.257)
    -wo^2= -1.0056rad/s
    wo^2 = 1.0056rad/s
    wo= 1.00265rad/s
    so at this point i took vt= rw vt= 3m x 1.00265rad/s vt = 3.008m/s

    I'm getting the answer wrong and I don't know why. Can anyone help me please?
  2. jcsd
  3. Dec 11, 2009 #2
    It could be that you are really 16 spaces away because of the direction you are required to spin in. Looks okay otherwise.
  4. Dec 11, 2009 #3
    thanks it seems to be the case one of the multiple choice answers is very close to this, but since it doesnt specify where on the wheel it is being pulled down from left side or right side this question is ambiguous
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