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Country Fair Spinning Wheel

  1. Mar 13, 2010 #1
    1. At a county fair there is a betting game that involves a spinning wheel. As the drawing shows, the wheel is set into rotational motion with the beginning of the angular section labeled "1" at the marker at the top of the wheel. The wheel then decelerates and eventually comes to a halt on one of the numbered sections. The wheel in the drawing is divided into twelve sections, each of which is an angle of 30˚. Determine the numbered section on which the wheel comes to a halt when the deceleration of the wheel has a magnitude of 0.355 rev/s2 and the initial angular velocity is (a) +1.61 rev/s and (b) +3.17 rev/s.

    http://edugen.wiley.com/edugen/courses/crs2216/art/qb/qu/c08/qu_8_32.gif



    2. Relevant equations
    I tried using equation [tex]\omega[/tex][tex]_{2}[/tex] = [tex]\omega[/tex][tex]^{2}_{o}[/tex] + 2[tex]\alpha[/tex][tex]\theta[/tex]


    3. The attempt at a solution
    I used 0 is omega final, +1.61 (a), and +3.17(b) are omega initials, alpha is 0.355, and theta is my unknown.

    For part a, I got 3.65 revolutions and part B I got 14.15 revolutions. I do not know if this is correct so far, but I do not know how to convert from revolutions to degrees.
     
  2. jcsd
  3. Mar 13, 2010 #2

    ideasrule

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    One revolution is 360 degrees.
     
  4. Mar 13, 2010 #3
    Thank you!

    However, I did not get the right answer when I converted my answers to degrees and then found the respective spot on the wheel.

    To do this, I found subtracted 360 from my answer until I got a number below 360. Then I when by 30's from left to right (counter-clockwise) on the wheel until I landed in the respective box.

    Any help with an error please?
     
  5. Mar 13, 2010 #4

    rl.bhat

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    ω = 2*π*f, where f is revolution per second. Try this one to find θ.
     
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