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Homework Help: Ferris wheel Rotational Kinematics Question

  1. Mar 25, 2010 #1
    1. The problem statement, all variables and given/known data
    The given information is:
    A Ferris wheel with a radius of 9.0 m rotates at a constant rate, completing one revolution every 37 s.
    Suppose the Ferris wheel begins to decelerate at the rate of 0.22 rad/s^2 when the passenger is at the top of the wheel.

    (1) Find the magnitude of the passenger's acceleration at that time.
    (2) Find the direction of the passenger's acceleration at that time.


    2. Relevant equations

    a_c=r*w^2
    a_t=r*alpha


    3. The attempt at a solution
    I tried to find the acceleration at the top using those formulas and couldn't get it. I thought I should try total acceleration using a= sqrt(a_c^2+a_t^2) and got 0.260 m/s, but that was no good either. Honestly what is really confusing me is what exactly I'm finding; I've calculated several types of acceleration during this homework so I'm not sure what just the "passenger's acceleration" is. Because of part two I'm guessing I need to incorporate vectors somehow, but I can't figure out what approach to take.

    Thanks for any help!
     
  2. jcsd
  3. Mar 25, 2010 #2

    Doc Al

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    Staff: Mentor

    That's correct.
    Show your calculations for a_c and a_t.
     
  4. Mar 25, 2010 #3
    I did:

    a= sqrt(a_c^2+a_t^2)

    a_c^2=r*w^2=9*.169815819^2=.259536711

    (I calculated w by doing 2pi/37seconds)

    a_t^2=r*theta=9*(-.22)^2=.4356

    (I used the given deceleration as theta, but I'm not sure if that works)

    so a= sqrt((.259536711^2)+(.4356^2))= .507

    As you can see I got a different answer on this new attempt. I suppose I'll try that, but I'm still not confident my decision to use -.22 as my acceleration works.
     
  5. Mar 25, 2010 #4

    Doc Al

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    I assume you meant this to be a_t, not a_t^2. Why did you square the angular acceleration? (Angular acceleration is usually represented by alpha, not theta.)
     
  6. Mar 25, 2010 #5
    Yes, I meant alpha. And I squared it because I was accidentally looking at the formula for centripetal acceleration when I was writing the tangential acceleration formula. Is that squared term the root of my problems?
     
  7. Mar 25, 2010 #6

    Doc Al

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    Yes, it looks like the main problem in that last attempt.
     
  8. Mar 25, 2010 #7
    That was right! Thanks so much. If you wouldn't mind I still can't get part two. I know that a_c and a_t are perpendicular to each other, so I was thinking I should use trigonometry to find the direction of acceleration. I tried arctan(a_c/a_t), but that was no good.
     
  9. Mar 25, 2010 #8

    Doc Al

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    That would work just fine, but you need to know how they want you to describe the direction. With respect to the vertical or the horizontal? (The angle that you calculated is with respect to what?)
     
  10. Mar 25, 2010 #9
    "_____degrees below the direction of motion" is what the field looks like. I thought my trig function would work in that situation.
     
  11. Mar 25, 2010 #10

    Doc Al

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    Yes, it should work. Show the details. What did you put for a_c, a_t, and theta?
     
  12. Mar 25, 2010 #11
    I did "arctan theta=.259536711/1.98" and I got 7.47 for theta.
     
  13. Mar 25, 2010 #12
    180-arctan(a_c/a_t) might give you your answers.
     
  14. Mar 25, 2010 #13

    Doc Al

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    Looks good, but what's the direction of motion? Remember that the acceleration is negative.

    :wink:
     
  15. Mar 25, 2010 #14
    That was right! Thank you both very much. I really appreciate it! :D
     
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